Selected definitions and basic equations
Photon energy
Eph = hy = hv; v = 2py
Attenuation in second medium in TIR
1>2
2pn2 n1 2 2
a2 =
c a b sin ui - 1 d
n2
lo
Photon momentum
h
= hk
l
pph =
Photon flux Φph and irradiance (intensity)
∆Nph
Photons crossing area A in time ∆t
Φph =
=
A∆t
A∆t
I = hyΦph
Propagation constant (wave vector)
2p
k =
l
Phase velocity
v
c
c
v = ly = ; v = =
n
k
1er
1
= veoer E 2o
2
Snell’s law and the Brewster angle
n2
n2
n1 sin ui = n2 sin ut; sin uc = ; tan up =
n1
n1
Phase change in total internal reflection (TIR)
tan 1 12f#2 =
2 1>2
[sin ui - n ]
cos ui
tan 112 (f // + p)2 =
(n1 + n2)2
Fabry–Perot cavity
c
ym = m a b = myf, m = 1, 2, 3, c
2L
yf
F
; F =
pR1>2
1 - R
I(u) = I(0)sinc2(b); b =
1
(ka sin u)
2
Airy disk, angular radius, divergence
l
sin uo = 1.22
D
Divergence = 2uo ≈ 2 * 1.22
l
D
Diffraction grating
d( sin um - sin ui) = ml; m = 0, {1, {2, c
V-number, normalized frequency
2pa 2
2pa
V =
(n1 - n22)1>2 ; V =
NA
l
l
c
Ex = vBy = By
n
Poynting vector and irradiance
2
4n1n2
Single slit diffraction
Electric and magnetic fields
S = v eoer E * B ; I = Saverage
T = T# = T// =
dym =
Changes in wavelength and frequency
dl
dy
l2
c
= - ; dl = - dy = - 2 dy
y
c
l
y
Group velocity
dv
vg =
dk
Group index
c
dn
vg(medium) =
; Ng = n - lo
Ng
dlo
2
Reflectance, transmittance (normal incidence)
n1 - n2 2
R = R# = R// = a
b ;
n1 + n2
;n =
n2
n1
[sin 2ui - n2 ]1>2
n2 cos ui
Normalized index difference
∆ = (n1 - n2)>n1
Acceptance angle and numerical aperture (NA)
2amax ; sin amax =
(n21 - n22)1>2
NA
; sin amax =
n0
n0
Normalized propagation constant
(b>k)2 - n22
(b>k) - n2
b =
≈
n1 - n2
n21 - n22
b ≈ a1.1428 -
0.996 2
b for 1.5 6 V 6 2.5
V
Single mode waveguides
Planar waveguide: V 6 p>2
Step@index fiber: V 6 2.405
Mode field diameter
2w = 2a(0.65 + 1.619V -3>2 + 2.879V -6);
0.8 6 V 6 2.5
Dispersion in multimode step-index fiber
n1 - n2
n1 ∆
∆t
≈
=
c
c
L
Dispersion coefficient
∆t>L = Spread in group delay per unit length
= D∆l
∆t
D =
L∆l
Chromatic dispersion
∆t
= 0 Dm + Dw + Dp 0 ∆l
L
Maximum RTZ bit rate
0.25
B ≈
s
Attenuation in optical fibers
Pin
1
adB = 10 log a
b = 4.34a
L
Pout
where a is the attenuation coefficient.
Optical gain coefficient
g(y) = sem(y)N2 - sab(y)N1
Optical gain
G = exp(gL)
Threshold gain in lasers
gth = as +
1
1
ln a
b = at
2L
R 1R 2
Photon cavity lifetime
tph ≈ n>cat
Bandgap light and wavelength
1.24
lg(om) =
Eg(eV)
Responsivity of a photodetector
R =
Iph
Photocurrent (A)
=
Incident optical power (W)
Po
External quantum efficiency of a photodetector
he =
Iph >e
Po >hy
Phase change between e- and o-waves
2p
f =
(n - no)L
l e
Arthur L. Schawlow is adjusting a ruby optical maser during an
­experiment at Bell Labs, while C.G.B. Garrett prepares to photograph the maser flash. In 1981, Arthur Schawlow shared the
Nobel Prize in Physics for his “contribution to the development
of laser spectroscopy.” (Reprinted with permission of AlcatelLucent USA Inc.)
The patent for the invention of the laser by Charles H. Townes and Arthur L. Schawlow in 1960 (Reprinted with permission of Alcatel-Lucent USA Inc.). This laser patent was later bitterly disputed for almost three decades in the so-called
“laser patent wars” by Gordon Gould, an American physicist, and his designated agents. Gordon Gould eventually received
the U.S. patent for optical pumping of the laser in 1977 inasmuch as the original laser patent did not detail such a pumping procedure. In 1987 he also received a patent for the gas discharge laser, thereby winning his 30-year patent war. His
original notebook even contained the word “laser.” (See “Winning the laser-patent war”, Jeff Hecht, Laser Focus World,
December 1994, pp. 49–51).
Second Edition
Optoelectronics
and Photonics:
Principles and Practices
S.O. Kasap
University of Saskatchewan
Canada
International Edition Contributions by
Ravindra Kumar Sinha
Delhi Technological University
India
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
10
Contents
1.11 Multiple Interference and Optical Resonators 69
Example 1.11.1 R
esonator modes and spectral width of a semiconductor
Fabry–Perot cavity 73
1.12 Diffraction Principles 74
A. Fraunhofer Diffraction 74
Example 1.12.1 Resolving power of imaging systems
79
B. Diffraction Grating 80
Example 1.12.2 A reflection grating 83
Additional Topics 84
1.13 Interferometers 84
1.14 Thin Film Optics: Multiple Reflections in Thin Films 86
Example 1.14.1 Thin film optics
88
1.15 Multiple Reflections in Plates and Incoherent Waves 89
1.16 Scattering of Light 90
1.17 Photonic Crystals 92
Questions and Problems 98
Chapter 2 Dielectric Waveguides and Optical Fibers 111
2.1 Symmetric Planar Dielectric Slab Waveguide 111
A. Waveguide Condition 111
B. Single and Multimode Waveguides 116
C. TE and TM Modes 116
Example 2.1.1 Waveguide modes 117
Example 2.1.2 V-number and the number of modes
Example 2.1.3 Mode field width, 2wo 119
118
2.2 M
odal and Waveguide Dispersion in Planar
Waveguides 120
A. Waveguide Dispersion Diagram and Group Velocity 120
B. Intermodal Dispersion 121
C. Intramodal Dispersion 122
2.3 Step-Index Optical Fiber 123
A. Principles and Allowed Modes 123
Example 2.3.1 A multimode fiber 128
Example 2.3.2 A single-mode fiber 128
B. Mode Field Diameter 128
Example 2.3.3 Mode field diameter 129
C. Propagation Constant and Group Velocity 130
Example 2.3.4 Group velocity and delay
131
D. Modal Dispersion in Multimode Step-Index Fibers 132
Example 2.3.5 A multimode fiber and dispersion
132
16
Contents
Example 5.12.2 Noise of an ideal photodetector
Example 5.12.3 SNR of a receiver 429
428
B. Avalanche Noise in the APD 430
Example 5.12.4 Noise in an APD
430
5.13 Image Sensors 431
A. Basic Principles 431
B. Active Matrix Array and CMOS Image Sensors 433
C. Charge-Coupled Devices 435
Additional Topics 437
5.14 Photovoltaic Devices: Solar Cells 437
A. Basic Principles 437
B. Operating Current and Voltage and Fill Factor 439
C. Equivalent Circuit of a Solar Cell 440
D. Solar Cell Structures and Efficiencies 442
Example 5.14.1 Solar cell driving a load 444
Example 5.14.2 Open circuit voltage and short circuit current 445
Questions and Problems 445
Chapter 6 Polarization and Modulation of Light
457
6.1 Polarization 457
A. State of Polarization 457
Example 6.1.1 Elliptical and circular polarization 460
B. Malus’s Law 460
6.2 L
ight Propagation in an Anisotropic Medium:
Birefringence 461
A. Optical Anisotropy 461
B. Uniaxial Crystals and Fresnel’s Optical Indicatrix 463
C. Birefringence of Calcite 466
D. Dichroism 467
6.3 Birefringent Optical Devices 468
A. Retarding Plates 468
Example 6.3.1 Quartz-half wave plate 469
Example 6.3.2 Circular polarization from linear polarization 470
B. Soleil–Babinet Compensator 470
C. Birefringent Prisms 471
6.4 Optical Activity and Circular Birefringence 472
6.5 Liquid Crystal Displays 474
6.6 Electro-Optic Effects 478
A. Definitions 478
Contents
B. Pockels Effect 479
Example 6.6.1 Pockels Cell Modulator 484
C. Kerr Effect 484
Example 6.6.2 Kerr Effect Modulator 486
6.7 Integrated Optical Modulators 486
A. Phase and Polarization Modulation 486
B. Mach–Zehnder Modulator 487
C. Coupled Waveguide Modulators 489
Example 6.7.1 Modulated Directional Coupler 492
6.8 Acousto-Optic Modulator 492
A. Photoelastic Effect and Principles 492
B. Acousto-Optic Modulators 494
Example 6.8.1 AO Modulator 499
6.9 Faraday Rotation and Optical Isolators 499
Example 6.9.1 Faraday rotation 500
6.10 Nonlinear Optics and Second Harmonic Generation 501
Additional Topics 505
6.11 Jones Vectors 505
Questions and Problems 506
Appendices
Appendix A­ Gaussian Distribution 514
Appendix B
Solid Angles 516
Appendix C
Basic Radiometry and Photometry 518
Appendix D­ Useful Mathematical Formulae 521
Appendix E­ Notation and Abbreviations 523
Index 535
CMOS image sensors with wide dynamic range. (Courtesy of
New Imaging Technologies (NIT), France)
17
—Sir William Henry Bragg1
Augustin Jean Fresnel (1788–1827) was a French physicist and a civil engineer for the French ­government
who was one of the principal proponents of the wave theory of light. He made a number of distinct contributions to optics including the well-known Fresnel lens that was used in lighthouses in the nineteenth
century. He fell out with Napoleon in 1815 and was subsequently put under house arrest until the end
of Napoleon’s reign. During his enforced leisure time he ­formulated his wave ideas of light into a mathematical theory. (© INTERFOTO/Alamy.)
—Attributed to Augustin Fresnel
1
As quoted in Alan Mackay, Dictionary of Scientific Quotes, 2nd Edition (Institute of Physics Publishing, Bristol, 1991), p. 37.
1.1 • Light Waves in a Homogeneous Medium
25
Figure 1.6 (a) Gaussian beam
definitions. The region zo is called
the Rayleigh range and also the
depth of focus. (b) Comparison of
a real beam with M 2 7 1 with a
Gaussian beam with M 2 = 1 and
the same waist 2wo.
Far away from the Rayleigh range, for z W zo, in the far-field region, the beam width
­increases linearly with z, that is,
2w ≈ (2wo)
z
zo
(1.1.9b)
Gaussian
beam
width at
distance z
Notice that the product of the beam radius wo (half the beam waist) and half the divergence angle u from Eq. (1.1.7) is given by wou = l>p, that is, it depends only on the wavelength and is a well-defined constant for a given wavelength. The product wou is called the beam
­parameter product.
The Gaussian beam concept is so useful in photonics that a special quantity, called the
M2-factor, has been introduced to compare a given laser beam to an ideal Gaussian beam. The
M2 factor measures the deviation of the real laser beam from the Gaussian characteristics, in
which M 2 = 1 for an ideal (theoretical) Gaussian beam shape. Suppose that 2ur and 2wor are the
divergence and waist, respectively, of the real laser beam, and 2u and 2wo are those for the ideal
Gaussian. The M2 factor is defined by6
M2 =
wor ur
wor ur
=
wou
(l>p)
(1.1.10)
where we have used wou = l>p for an ideal Gaussian beam.
According to Eq. (1.1.10), M2 is the ratio of the beam parameter product of the real
beam to that of a Gaussian beam, and hence M2 gauges the beam quality of the laser beam.
For many ­lasers, M2 is greater than unity, and can be as high as 10–30 in multimode lasers.
6
Some authors define M, instead of M2, as M = ur >u = wor >wo. In addition, the reader should not be too concerned with
the terms “multimode” and “single mode” at this point, except that they represent the types of radiation that is emitted
from lasers.
M2 factor
definition
28
Chapter 1 • Wave Nature of Light
Table 1.1 Low-frequency (LF) relative permittivity Er (LF) and refractive index n
Material
Er(LF)
[Er(LF)]1,2
Si
Diamond
GaAs
SiO2
Water
11.9
5.7
13.1
3.84
80
3.44
2.39
3.62
2.00
8.9
Comment
n (at L)
3.45 (at 2.15 om)
2.41 (at 590 nm)
3.30 (at 5 om)
1.46 (at 600 nm)
1.33 (at 600 nm)
Electronic bond polarization up to optical frequencies
Electronic bond polarization up to UV light
Ionic polarization contributes to er(LF)
Ionic polarization contributes to er(LF)
Dipolar polarization contributes to er(LF), which
is large
mechanisms operate at these frequencies.9 At low frequencies all polarization mechanisms present can contribute to er, whereas at optical frequencies only the electronic polarization can respond to the oscillating field. Table 1.1 lists the relative permittivity er(LF) at low frequencies
(e.g., 60 Hz or 1 kHz as would be measured, for example, using a capacitance bridge in the
laboratory) for various materials. It then compares 3er(LF)4 1>2 with n.
For silicon and diamond there is an excellent agreement between 3er(LF)4 1>2 and n. Both
are covalent solids in which electronic polarization (electronic bond polarization) is the only
polarization mechanism at low and high frequencies. Electronic polarization involves the displacement of light electrons with respect to positive ions of the crystal. This process can readily
respond to the field oscillations up to optical or even ultraviolet frequencies.
For GaAs and SiO2 3 er(LF)4 1>2 is larger than n because at low frequencies both of these solids possess a degree of ionic polarization. The bonding is not totally covalent and there is a degree
of ionic bonding that contributes to polarization at frequencies below far-infrared wavelengths.
In the case of water, the er(LF) is dominated by orientational or dipolar polarization, which
is far too sluggish to respond to high-frequency oscillations of the field at optical frequencies.
It is instructive to consider what factors affect n. The relative permittivity depends on
the ­polarizability a per molecule (or atom) in the solid. (a is defined as the induced electric
dipole moment per unit applied field.) The simplest and approximate expression for the relative
permittivity is
er ≈ 1 +
Cauchy
short form
dispersion
equation
Na
eo
in which N is the number of molecules per unit volume. Both the atomic concentration, or density, and polarizability therefore increase n. For example, glasses of given type but with greater
density tend to have higher n.
The frequency or wavelength dependence of er and hence n is called the dispersion relation,
or simply dispersion. There are various theoretical and empirical models that describe the n vs. l
behavior. The Cauchy dispersion equation in its simplest form is given by10
n = A +
B
C
+ 4
l2
l
(1.2.3)
9
Chapters 7 and 9 in Principles of Electronic Materials and Devices, 3rd Edition, S. O. Kasap (McGraw-Hill, 2006)
provides a semiquantitative description of the frequency dependence of er and hence the wavelength dependence of n.
10
Dispersion relations like the one in Eq. (1.2.3) are always in terms of the free-space wavelength l. (It does not make
sense to give them in terms of the actual wavelength in the medium.)
1.2 • Refractive Index and Dispersion
29
Table 1.2 Sellmeier and Cauchy coefficients
Sellmeier
L1 (,m)
L2 (,m)
L3 (,m)
A1
A2
A3
SiO2 (fused
silica)
86.5%SiO213.5%GeO2
0.696749
0.408218
0.890815
0.0690660
0.115662
9.900559
0.711040
0.451885
0.704048
0.0642700
0.129408
9.425478
GeO2
Sapphire
Diamond
0.80686642
1.023798
0.3306
0.71815848
1.058264
4.3356
0.85416831
5.280792
–
0.068972606
0.0614482
0.1750
0.15396605
0.110700
0.1060
11.841931
17.92656
–
Cauchy
Range of
hv (eV)
n0
n2 (eV–2)
n4 (eV–4)
Diamond
0.05-5.47
- 1.07 * 10
Silicon
0.002-1.08
Germanium
0.002-0.75
n–2 (eV2)
2.378
8.01 * 10
-3
1.04 * 10-4
- 2.04 * 10-8
3.4189
8.15 * 10-2
1.25 * 10-2
- 1.0 * 10-8
4.003
2.2 * 10-1
1.4 * 10-1
-5
Source: Sellmeier coefficients combined from various sources. Cauchy coefficients from D. Y. Smith et al., J. Phys. CM, 13, 3883, 2001.
where A, B, and C are material-specific constants. A more general Cauchy dispersion relation is
of the form
n = n-2(hy)-2 + n0 + n2(hy)2 + n4(hy)4
(1.2.4)
where hy is the photon energy, and n0, n-2, n2, and n4 are constants; values for diamond, Si, and
Ge are listed in Table 1.2. The general Cauchy equation is usually applicable over a wide photon
energy range.
Another useful dispersion relation that has been widely used, especially in optical fibers,
is the Sellmeier equation given by
n2 = 1 +
A1l2
l2 - l21
+
A2l2
l2 - l22
+
A3l2
l2 - l23
(1.2.5)
where A1, A2, A3 and l1, l2, l3 are constants, called Sellmeier coefficients.11 Equation (1.2.5)
turns out to be quite a useful semi-empirical expression for calculating n at various wavelengths
if the Sellmeier coefficients are known. Higher terms involving A4 and higher A coefficients can
generally be neglected in representing n vs. l behavior over typical wavelengths of interest. For
example, for diamond, we only need the A1 and A2 terms. The Sellmeier coefficients are listed in
various optical data handbooks.
Example 1.2.1 Sellmeier equation and diamond
Using the Sellmeier coefficients for diamond in Table 1.2, calculate its refractive index at 610 nm (red light)
and compare with the experimental quoted value of 2.415 to three decimal places.
11
This is also known as the Sellmeier-Herzberger formula.
Cauchy
dispersion
equation
in photon
energy
Sellmeier
equation
40
Chapter 1 • Wave Nature of Light
When n1 7 n2 then obviously the transmitted angle is greater than the incidence angle as
apparent in Figure 1.11. When the refraction angle ut reaches 90°, the incidence angle is called
the critical angle uc, which is given by
Total
internal
reflection
(TIR)
sin uc =
n2
n1
(1.5.2)
When the incidence angle ui exceeds uc then there is no transmitted wave but only a ­reflected
wave. The latter phenomenon is called total internal reflection (TIR). The effect of ­increasing the
incidence angle is shown in Figure 1.12. It is the TIR phenomenon that leads to the propagation
of waves in a dielectric medium surrounded by a medium of smaller refractive index as shown in
Chapter 2. Although Snell’s law for ui 7 uc shows that sin ut 7 1 and hence ut is an “imaginary”
angle of refraction, there is however a wave called the evanescent wave, whose amplitude decays
exponentially with distance into the second medium as discussed below. The wave exists only in
the interface region from which the reflected wave emerges (not outside).
Snell’s law can also be viewed as the k-vector of light parallel to the interface being continuous through the interface, that is, having the same value on both sides of the interface. In
­medium n1, ki parallel to the interface is ki sin ui or kn1 sin ui, where ki = kn1, and k is the magnitude of the wave vector in free space. In medium n2, kt parallel to the interface is kt sin ut or
kn2 sin ut. If k’s component tangential to the interface remains constant, kn1 sin ui = kn2 sin ut,
then we obtain Snell’s law in Eq. (1.5.1). Put differently, Snell’s law is equivalent to
Snell’s
Law
n sin u = constant through an interface between different media
(1.5.3)
Snell’s law of refraction and TIR play a very important role in many optoelectronic and
photonic devices. A prism is a transparent optical component that can deflect a light beam as
illustrated in Figure 1.13. There are two basic types of prism. In a refracting prism, the light
­deflection is caused by refractions whereas in a reflecting prism it is caused by one or more
TIRs. (Some prisms such as composite prisms need both refraction and TIR to achieve their
­desired deflection.) The deflection d depends not only on the incidence angle of the light beam on
the prism, the prism material (n), and geometry, but also on the wavelength and the p­ olarization
state of the incident light. The reason is that the refractive index n of the prism material normally
depends on the wavelength, and further, for certain materials (e.g., quartz, calcite), it depends on
the polarization state (direction of the electric field) of light as well.
Figure 1.12 Light wave travelling in a more dense medium strikes a less dense medium. Depending on the
incidence angle with respect to uc, which is determined by the ratio of the refractive indices, the wave may be
transmitted (refracted) or reflected. (a) ui 6 uc (b) ui = uc (c) ui 7 uc and total internal reflection. (Wavefronts
are only indicated in (a).)
44
Chapter 1 • Wave Nature of Light
and
Transmission
coefficient
t# =
Eto,#
Eio,#
=
2cos ui
cos ui + 3n2 - sin2 ui4 1>2
(1.6.6b)
There are corresponding coefficients for the E// fields with corresponding reflection and
transmission coefficients, r// and t//.
Reflection
coefficient
r// =
Transmission
coefficient
t// =
Ero, //
Eio, //
Eto, //
Eio, //
=
=
3n2
3n2
- sin2 ui4 1/2 - n2 cos ui
- sin2 ui4 1>2 + n2 cos ui
2n cos ui
n cos ui + 3n2 - sin2 ui4 1>2
2
(1.6.7a)
(1.6.7b)
Further, the coefficients above are related by
Transmission
coefficient
r// + nt// = 1 and r# + 1 = t#
(1.6.8)
The significance of these equations is that they allow the amplitudes and phases of the
reflected and transmitted waves to be determined from the coefficients r›, r//, t//, and t›. For
convenience we take Eio to be a real number so that phase angles of r› and t› correspond to the
phase changes measured with respect to the incident wave. For example, if r› is a complex quantity then we can write this as r# = r# exp(-jf #) in which r› and f # represent the relative
amplitude and phase of the reflected wave with respect to the incident wave for the field perpendicular to the plane of incidence. Of course, when r› is a real quantity, then a positive number
represents no phase shift and a negative number is a phase shift of 180° (or p). As with all waves,
a negative sign corresponds to a 180° phase shift. Complex coefficients can be obtained only from
Fresnel’s equations if the terms under the square roots become negative and this can happen only
when n 6 1 (or n1 7 n2), and also when ui 7 uc, the critical angle. Thus, phase changes other
than 0 or 180° occur only when there is total internal reflection. Fresnel’s equations for normal
incidence are greatly simplified. Putting ui = 0 into Eqs. (1.6.6) and (1.6.7) we find
Normal
incidence
r// = r# =
n1 - n2
n1 + n2
and t// = t# =
2n1
n1 + n2
(1.6.9)
Figure 1.16 (a) shows how the magnitudes of the reflection coefficients, r› and r// ,
vary with the incidence angle ui for a light wave traveling from a more dense medium, n1 = 1.44,
to a less dense medium, n2 = 1.00, as predicted by Fresnel’s equations. Figure 1.16 (b) shows
the changes in the phase of the reflected wave, f# and f //, with ui. The critical angle uc as
­determined from sin uc = n2 >n1 in this case is 44°. It is clear that for incidence close to normal
(small ui), there is no phase change in the reflected wave. The reflection coefficient in Eq. (1.6.9)
is a positive quantity for n1 7 n2, which means that the reflected wave suffers no phase change.
This is confirmed by f # and f // in Figure 1.16 (b). As the incidence angle increases, eventually r// becomes zero at an angle of about 35°. We can find this special incidence angle, labeled
as up, by solving the Fresnel equation (1.6.7a) for r// = 0. The field in the reflected wave is
then always perpendicular to the plane of incidence and hence well-defined as illustrated in
1.6 • Fresnel’s Equations
45
Figure 1.16 Internal reflection: (a) Magnitude of the reflection coefficients r// and r› vs. angle of incidence
ui for n1 = 1.44 and n2 = 1.00. The critical angle is 44°. (b) The corresponding phase changes f// and f# vs.
incidence angle ui.
Figure 1.17. This special angle is called the polarization angle or Brewster’s19 angle and
from Eq. (1.6.7a) is given by
tan up =
n2
n1
(1.6.10)
The reflected wave is then said to be linearly polarized because it contains electric field
oscillations that are contained within a well-defined plane, which is perpendicular to the plane
of incidence and also to the direction of propagation. Electric field oscillations in unpolarized
light, on the other hand, can be in any one of infinite number of directions that are perpendicular
Figure 1.17 At the Brewster angle
of incidence ui = up, the reflected light
contains only field oscillations normal
to the plane of incidence (paper).
19
After Sir David Brewster (1781–1868), a Scottish physicist who was educated in theology at the University of
Edinburgh in Scotland. He became interested in the polarization properties of light from 1799 onwards, and reported
some of his experiments in scientific journals, including the Philosophical Transactions of London.
Brewster’s
polarization
angle
46
Chapter 1 • Wave Nature of Light
to the direction of propagation. In linearly polarized light, however, the field oscillations are
contained within a well-defined plane. Light emitted from many light sources, such as a tungsten
light bulb or an LED diode, is unpolarized.20 Unpolarized light can be viewed as a stream or
collection of EM waves whose fields are randomly oriented in a direction that is perpendicular
to the direction of light propagation.
For incidence angles greater than up but smaller than uc, Fresnel’s equation (1.6.7a) gives
a negative number for r//, which indicates a phase shift of 180° as shown in f // in Figure 1.16
(b). The magnitude of both r// and r› increases with ui as illustrated in Figure 1.16 (a). At the
critical angle and beyond (past 44° in Figure 1.16), that is, when ui Ú uc, the magnitudes of
both r// and r› go to unity so that the reflected wave has the same amplitude as the incident
wave. The incident wave has suffered total internal reflection, TIR. When ui 7 uc, in the
presence of TIR, both Eqs. (1.6.6) and (1.6.7) are complex quantities because then sin ui 7 n
and the terms under the square roots become negative. The reflection coefficients become complex quantities of the type r# = 1 # exp (-jf #) and r// = 1 # exp (-jf //) with the phase angles
f # and f // being other than zero or 180°. The reflected wave therefore suffers phase changes,
f # and f //, in the components E› and E//. These phase changes depend on the incidence angle,
as illustrated in Figure 1.16 (b), and on n1 and n2.
Examination of Eq. (1.6.6) for r› shows that for ui 7 uc, we have |r#| = 1, but the phase
change f # is given by
Phase
change
in TIR
tan 1 12 f #2 =
3sin2 ui
- n2 41>2
cos ui
(1.6.11)
For the E// component, the phase change f // is given by
Phase
change
in TIR
tan 112 f // +
1
2
p2 =
3sin2 ui
- n2 41>2
n2 cos ui
(1.6.12)
We can summarize that in internal reflection (n1 7 n2), the amplitude of the reflected
wave from TIR is equal to the amplitude of the incident wave but its phase has shifted by an
amount determined by Eqs. (1.6.11) and (1.6.12). The fact that f // has an additional p shift that
makes f // negative for ui 7 uc is due to the choice for the direction of the reflected optical field
Er, // in Figure 1.15. This p shift can be ignored if we by simply invert Er, //. (In many books
Eq. (1.6.12) is written without the p-shift.)
The reflection coefficients in Figure 1.16 considered the case in which n1 7 n2. When
light approaches the boundary from the higher index side, that is n1 7 n2, the reflection is said
to be internal reflection and at normal incidence there is no phase change. On the other hand,
if light approaches the boundary from the lower index side, that is n1 6 n2, then it is called
external reflection. Thus in external reflection light becomes reflected by the surface of an
optically denser (higher refractive index) medium. There is an important difference between
the two. Figure 1.18 shows how the reflection coefficients r› and r// depend on the incidence
angle ui for external reflection (n1 = 1 and n2 = 1.44). At normal incidence, both coefficients
are negative, which means that in external reflection at normal incidence there is a phase shift
of 180°. Further, r// goes through zero at the Brewster angle up given by Eq. (1.6.10). At this
20
Even light from a tungsten light bulb or an LED has some polarization but this is usually negligibly small.
1.6 • Fresnel’s Equations
47
Figure 1.18 External reflection coefficients
r// and r› vs. angle of incidence ui for n1 = 1.00
and n2 = 1.44.
angle of incidence, the reflected wave is polarized in the E› component only. Transmitted
light in both internal reflection (when ui 6 uc) and external reflection does not experience a
phase shift.
What happens to the transmitted wave when ui 7 uc ? According to the boundary conditions, there must still be an electric field in medium 2; otherwise, the boundary conditions
cannot be satisfied. When ui 7 uc, the field in medium 2 is a wave that travels near the
­surface of the boundary along the z direction as shown in Figure 1.19. The wave is called an
evanescent wave and advances along z with its field decreasing as we move into medium 2;
that is,
Et,#(y, z, t) ∝ e-a2y exp j(vt - kizz)
(1.6.13)
Evanescent
wave
in which kiz = ki sin ui is the wave vector of the incident wave along the z-axis, and a2 is an
a­ ttenuation coefficient for the electric field penetrating into medium 2,21
a2 =
1>2
2pn2 n1 2 2
c a b sin ui - 1 d
n2
lo
(1.6.14)
in which l is the free-space wavelength. According to Eq. (1.6.13), the evanescent wave
travels along z and has an amplitude that decays exponentially as we move from the boundary into medium 2 (along y). The field of the evanescent wave is e–1 in medium 2 when
y = 1>a2 = d, which is called the penetration depth. It is not difficult to show that the
evanescent wave is correctly predicted by Snell’s law when ui 7 uc. The evanescent wave
propagates along the boundary (along z) with the same speed as the z-component velocity
of the incident and reflected waves. In Eqs. (1.6.1) and (1.6.2) we had assumed that the
incident and reflected waves were plane waves, that is, of infinite extent. If we were to
extend the plane wavefronts on the reflected wave, these would cut the boundary as shown
in Figure 1.19. The evanescent wave traveling along z can be thought of arising from these
plane wavefronts at the boundary as in Figure 1.19. (Evanescent wave is important in light
propagation in optical waveguides such as optical fibers.) If the incident wave is a narrow
21
Normally the term attenuation coefficient refers to the attenuation of the irradiance but in this case it refers to the
electric field.
Attenuation
of evanescent
wave
48
Chapter 1 • Wave Nature of Light
Figure 1.19 When ui 7 uc for a plane
wave that is reflected, there is an evanescent
wave at the boundary whose magnitude
decays into the n2-medium.
beam of light (e.g., from a laser pointer) then the reflected beam would have the same crosssection. There would still be an evanescent wave at the boundary, but it would exist only
within the cross-sectional area of the reflected beam at the boundary.
B. Intensity, Reflectance, and Transmittance
It is frequently necessary to calculate the intensity or irradiance22 of the reflected and transmitted
waves when light traveling in a medium of index n1 is incident at a boundary where the refractive index changes to n2. In some cases we are simply interested in normal incidence where
ui = 0°. For example, in laser diodes light is reflected from the ends of an optical cavity where
there is a change in the refractive index.
For a light wave traveling with a velocity v in a medium with relative permittivity er, the
light intensity I is defined in terms of the electric field amplitude Eo as
Light
intensity or
irradiance
1
ve e E 2
(1.6.15)
2 r o o
Here 12 ereoE 2o represents the energy in the field per unit volume. When multiplied by the
velocity v it gives the rate at which energy is transferred through a unit area. Since v = c>n and
er = n2 the intensity is proportional to nE 2o.
Reflectance R measures the intensity of the reflected light with respect to that of the incident light and can be defined separately for electric field components parallel and perpendicular
to the plane of incidence. The reflectances R› and R// are defined by
I =
R# =
Reflec­tances
0 Ero,# 0 2
0 Ero,// 0 2
2
0
0
=
r
and
R
=
= 0 r// 0 2
#
//
0 Eio,# 0 2
0 Eio,// 0 2
(1.6.16)
Although the reflection coefficients can be complex numbers that can represent phase
changes, reflectances are necessarily real numbers representing intensity changes. Magnitude of
a complex number is defined in terms of its product with its complex conjugate. For example,
when Ero, // is a complex number then
Ero,// 2 = (Ero,//)(Ero,//)*
in which (Ero, //)* is the complex conjugate of (Ero, //).
22
Strictly the terms intensity and irradiance are not the same as mentioned in footnote 15.
1.6 • Fresnel’s Equations
(b)
If light is traveling from glass to air, what is the reflection coefficient and the intensity of the
­reflected light?
Solution
(a)
The light travels in air and becomes partially reflected at the surface of the glass that corresponds to
external reflection. Thus n1 = 1 and n2 = 1.5. Then
r// = r# =
n1 - n2
1 - 1.5
=
= -0.2
n1 + n2
1 + 1.5
This is negative, which means that there is a 180° phase shift. The reflectance (R), which gives the
fractional reflected power, is
R = r 2// = 0.04 or 4,.
(b) The light travels in glass and becomes partially reflected at the glass–air interface that corresponds
to internal reflection. Thus n1 = 1.5 and n2 = 1. Then
r// = r# =
n1 - n2
1.5 - 1
=
= 0.2
n1 + n2
1.5 + 1
There is no phase shift. The reflectance is again 0.04 or 4%. In both cases (a) and (b), the amount of
reflected light is the same.
Example 1.6.3
Reflection and transmission at the Brewster angle
A light beam traveling in air is incident on a glass plate of refractive index 1.50. What is the Brewster or
polarization angle? What are the relative intensities of the reflected and transmitted light for the polarization
perpendicular and parallel to the plane of incidence at the Brestwer angle of incidence?
Solution
Light is traveling in air and is incident on the glass surface at the polarization angle up. Here n1 = 1, n2 = 1.5,
n = n2 >n1 = 1.5, and tan up = (n2 >n1) = 1.5 so that up = 56.31°. We now use Fresnel’s equations to find
the reflected and transmitted amplitudes. For the perpendicular polarization, from Eq. (1.6.6a),
r# =
cos (56.31°) - 31.52 - sin2 (56.31°)4 1>2
cos (56.31°) + 31.52 - sin2 (56.31°)4 1>2
= -0.385
On the other hand, r// = 0. The reflectances R# = r# 2 = 0.148 and R// = r// 2 = 0 so that the
reflected light has no parallel polarization in the plane of incidence. Notice the negative sign in r›, which
indicates a phase change of p.
From Eqs. (1.6.6b) and (1.7.7b), the transmission coefficients are
t# =
and
t// =
2cos (56.31°)
cos (56.31°) + 31.52 - sin2 (56.31°)4 1>2
= 0.615
2(1.5) cos (56.31°)
(1.5)2 cos (56.31°) + 31.52 - sin2 (56.31°)4 1>2
= 0.667
53
54
Chapter 1 • Wave Nature of Light
Notice that r// + nt// = 1 and r# + 1 = t#, as we expect. To find the transmittance for each polarization, we need the refraction angle ut. From Snell’s law, n1 sin ui = n2 sin ut, that is (1) sin (56.31°) = (1.5) sin ut,
we find ut = 33.69°. Then, from Eq. (1.6.20),
T# = c
(1.5) cos (33.69°)
(1.5) cos (33.69°)
d (0.615)2 = 0.852 and T// = c
d (0.667)2 = 1
(1) cos (56.31°)
(1) cos (56.31°)
Clearly, light with polarization parallel to the plane of incidence has greater intensity. Note that
R + T = 1 for both polarizations.
If we were to reflect light from a glass plate, keeping the angle of incidence at 56.3°, then the reflected
light will be polarized with an electric field component perpendicular to the plane of incidence. The transmitted light will have the field greater in the plane of incidence; that is, it will be partially polarized. By using
a stack of glass plates one can increase the polarization of the transmitted light. (This type of pile-of-plates
polarizer was invented by Dominique F. J. Arago in 1812.)
1.7 Antireflection Coatings and Dielectric Mirrors
Fresnel equations are routinely used in a number of applications in optoelectronics to design and
fabricate optical coatings, that is, thin films, to reduce reflections and glare, and also in various
components such as dielectric mirrors and filters. Section 1.14 on thin films optics provides a
good example of their application for thin film coatings; but in this section we consider two practical applications in optoelectronics: antireflection (AR) coatings and dielectric mirrors.
A. Antireflection Coatings on Photodetectors and Solar Cells
When light is incident on the surface of a semiconductor, it becomes partially reflected. Partial
reflection is an important consideration in solar cells where transmitted light energy into the
semiconductor device is converted to electrical energy. The refractive index of Si is about 3.5
at wavelengths around 600–800 nm. Thus, the reflectance with n1(air) = 1 and n2(Si) ≈ 3.5 is
R = a
n1 - n2 2
1 - 3.5 2
b = a
b = 0.309
n1 + n2
1 + 3.5
This means that 30% of the light is reflected and is not available for conversion to electrical energy; a considerable reduction in the efficiency of the solar cell.
However, we can coat the surface of the semiconductor device with a thin layer of a dielectric material, such as an a@Si1 - xNx:H (amorphous hydrogenated silicon nitride based on silicon
nitride, Si3N4, and x is typically 0.4–0.6), that has an intermediate refractive index. Figure 1.23
Figure 1.23 Illustration of how an antireflection (AR)
coating reduces the reflected light intensity. The thickness d
and the refractive index n2 of the antireflection coefficient
are such that the waves A and B have a phase difference of p
and hence interfere destructively. There is no reflection.
1.7 • Antireflection Coatings and Dielectric Mirrors
Figure 1.24 (a) Schematic illustration of the principle of the dielectric mirror with many high and low
refrac­tive index layers. Reflected waves A, B, C, D, and so on all interfere constructively if the layer thicknesses d1
and d2 are a quarter of a wavelength within the layer, that is d1 = l>4n1 and d2 = l>4n2, where l is the free-space
wavelength. The dielectric mirror is assumed to be coated on a substrate with an index n3. (b) The reflectance
of three different dielectric mirrors that have N = 10, n1 >n2 = 2.35>1.46; N = 10, n1 >n2 = 1.95>1.46; N = 6,
n1 >n2 = 1.95>1.46, n3 = 1.52. (Note: n(TiO2) = 2.35, n(Si3N4) = 1.95, n(SiO2) = 1.46.)
mirrors are widely used in photonics, for example, in solid state lasers such as the vertical cavity
surface emitting laser diode. Since the dielectric mirror has a periodic variation in the refractive
index (the period being d1 + d2), similar to a diffraction grating, it is sometimes referred to as a
Bragg ­reflector.24 It is left as an exercise to show that if we interchange the high and low layers,
n1 = nL and n2 = nH, we obtain the same result.
As shown in Figure 1.24 (b), with sufficient number of double layers, the reflectance is
almost unity over a band of wavelengths, ∆l. Conversely, the transmittance vanishes over the
same wavelength range ∆l. This wavelength range in which the transmittance vanishes is called
reflectance bandwidth; or the stop band for the transmitted light. As discussed in Section 1.17,
the dielectric mirror in Figure 1.24 (a) is a one-dimensional photonic crystal in which there is a
certain stop band within which there can be no propagation of waves along the z axis within the
multilayer dielectric structure in Figure 1.24 (a).
24
As we will see later, a periodic variation in the refractive index is actually a diffraction grating and is able to diffract or
reflect a wave in a certain direction if the periodicity is right. The dielectric mirror is also called a one-dimensional Bragg
grating structure. It is also a one-dimensional photonic crystal for large N.
57
58
Chapter 1 • Wave Nature of Light
Various dielectric mirrors, which are quarter
wave dielectric stacks on Pyrex or Zerodur
­substrates. (Courtesy of Newport.)
There are two general observations from the reflectance spectra in Figure 1.24 (b). The
reflectance R increases with N, the number of double layers used. R also increases with the
­refractive index ratio n1 >n2; or put differently, with the index contrast. For a large number of
layers, the bandwidth ∆l of the dielectric mirror increases with the index contrast. The maximum
reflectance RN for N pairs of layers is given by
Maximum
reflectance,
dielectric
mirror
RN = c
2N
n2N
1 - (n0 >n3)n2
n2N
1
+
(n0 >n3)n2N
2
2
d (1.7.3)
The bandwidth ∆l when 2N is large (for near-unity reflectance) is given by
n1 - n2
∆l
≈ (4>p) arcsin a
b (1.7.4)
lo
n1 + n2
Reflectance
bandwidth
which increases with the refractive index contrast as apparent from Figure 1.24 (b). If we interchange the low-high layers so that the high index layer is on the air side, there is very little change
in the reflectance or the bandwidth in a highly reflecting dielectric mirror.
Example 1.7.2 Dielectric mirror
Consider a dielectric mirror that has quarter wave layers consisting of Ta2O5 with nH = 1.78 and SiO2 with
nL = 1.55 both at 850 nm, the central wavelength at which the mirror reflects light. Suppose the substrate
is Pyrex glass with an index ns = 1.47 and the outside medium is air with n0 = 1. Calculate the maximum
reflectance of the mirror when the number N of double layers is 4 and 12. What would happen if you use
TiO2 with nH = 2.49, instead of Ta2O5? What is the bandwidth for these two dielectric stacks when they
are highly reflecting (with many pairs of layers)? Suppose we use a Si wafer as the substrate, what happens
to the maximum reflectance?
Solution
We use n0 = 1 for air, n1 = nH = 1.78, n2 = nL = 1.55, n3 = ns = 1.47, N = 4 in Eq. (1.7.3). Thus, for
four pairs of layers, the maximum reflectance R4 is
R4 = c
(1.78)2(4) - (1>1.47)(1.55)2(4)
(1.78)2(4) + (1>1.47)(1.55)
2
d = 0.40 or 40,
2(4)
If we repeat the calculation for 12 pairs of layers, we will find R12 = 90.6,.
1.8 • Absorption of Light and Complex Refractive Index
59
If we use TiO2 with n1 = nH = 2.49, we would find R4 = 94.0, and R12 = 100, (to two decimal
places). Obviously the refractive index contrast is important; with the TiO2-SiO2 stack we only need four
double layers to get roughly the same reflectance as from 12 pairs of layers of Ta2O5-SiO2. If we i­ nterchange
nH and nL in the 12-pair stack, that is, n1 = nL and n2 = nH, the Ta2O5-SiO2 reflectance falls to 80.8% but
the TiO2-SiO2 stack is unaffected since it is already reflecting nearly all the light.
We can only compare bandwidths ∆l for “infinite” stacks (those with R ≈ 100,). For the TiO2-SiO2
stack, Eq. (1.7.4) gives
∆l ≈ lo(4>p) arcsin a
n2 - n1
2.49 - 1.55
b = (850 nm)(4>p) arcsin a
b = 254 nm
n2 + n1
2.49 + 1.55
On the other hand, for the Ta2O5-SiO2 infinite stack, we get ∆l = 74.8 nm. As expected, ∆l is
­narrower for the smaller contrast stack.
If we change the substrate to a silicon wafer with n3 = ns = 3.50, we would find that the Ta2O5-SiO2
4-pair stack gives a reflectance of 68.5%, higher than before because the large index changes from nL to ns at
the substrate interface provides further reflections.
1.8 Absorption of Light and Complex Refractive Index
Generally when light propagates through a material it becomes attenuated in the direction of
propagation as illustrated in Figure 1.25. We distinguish between absorption and scattering both
of which gives rise to a loss of intensity in the regular direction of propagation. In absorption,
the loss in the power in the propagating electromagnetic wave is due to the conversion of light
energy to other forms of energy; for example, lattice vibrations (heat) during the polarization of
the molecules of the medium or during the local vibrations of impurity ions driven by the optical
field. The excitation of electrons from the valence band to the conduction band, or from impurities to the conduction band, in insulators and semiconductors would also absorb energy from the
propagating radiation. Further, free electrons inside a medium with a finite conductivity can be
drifted by the optical field in the radiation. As these electrons become scattered by lattice vibrations (or impurities) they will pass the energy they have acquired from the EM wave to lattice
vibrations. There are other examples as well. In all cases, some of energy from the propagating
light wave is absorbed and converted to other forms of energy.
On the other hand, scattering is a process by which the energy from a propagating EM wave
is redirected as secondary EM waves in various directions away from the original direction of propagation. (This is discussed later in this chapter) The attenuation coefficient a is defined as the
fractional decrease in the irradiance I of a wave per unit distance along the direction of propagation z
a = -
dI
Idz
Figure 1.25 Attenuation of a
traveling wave in a medium results
in the decay of its amplitude.
(1.8.1)
Definition
of attenuation
coefficient
60
Chapter 1 • Wave Nature of Light
When the irradiance decreases, dI>dz is negative, and the attenuation coefficient is a positive
­number. If the attenuation of the wave is due to absorption only, then a is the absorption coefficient.
It is instructive to consider what happens when a monochromatic light wave such as
Lossless
propagation
E = Eo exp j (vt - kz)
(1.8.2)
is propagating in a dielectric medium. The electric field E in Eq. (1.8.2) is either parallel to x or y
since propagation is along z. As the wave travels through the medium, the molecules become
polarized. This polarization effect is represented by the relative permittivity er of the medium.
If there were no losses in the polarization process, then the relative permittivity er would be
a real number and the corresponding refractive index n = e1>2
would also be a real number.
r
However, we know that there are always some losses in all polarization processes. For example,
when the ions of an ionic crystal are displaced from their equilibrium positions by an alternating electric field and made to oscillate, some of the energy from the electric field is coupled
and converted to lattice vibrations (called phonons). These losses are generally accounted
by describing the whole medium in terms of a complex relative permittivity (or dielectric
constant) er , that is,
Complex
dielectric
constant
er = er= - jer==
(1.8.3)
where the real part er= determines the polarization of the medium with losses ignored and the
imaginary part e″r describes the losses in the medium.25 For a lossless medium, obviously
er = er= . The loss e″r depends on the frequency of the wave and usually peaks at certain natural
(resonant) frequencies involved in the absorption process.
An EM wave that is traveling in a medium and experiencing attenuation due to absorption
can be generally described by a complex propagation constant k, that is,
Complex
propagation
constant
k = k = - jk″
(1.8.4)
where k′ and k″ are the real and imaginary parts. If we put Eq. (1.8.4) into Eq. (1.8.2) we will
find the following:
Attenuated
propagation
E = Eo exp (-k″z) exp j(vt - k′z)
(1.8.5)
The amplitude decays exponentially while the wave propagates along z. The real k′ part of
the complex propagation constant (wave vector) describes the propagation characteristics (e.g.,
phase velocity v = v>k =). The imaginary k″ part describes the rate of attenuation along z. The
irradiance I at any point along z is
I ∝ E 2 ∝ exp (-2k″z)
so that the rate of change in the irradiance with distance is
Imaginary
part k′
dI>dz = -2k″I
(1.8.6)
where the negative sign represents attenuation. Clearly a = 2k″
25
See, for example, S. O. Kasap, Principles of Electronic Materials and Devices, 3rd Edition (McGraw-Hill, 2006), Ch. 7.
Further, some books use er = er= + jer== instead of Eq. (1.8.3), with a positive imaginary part, but the latter normally refers
to an applied field that has a time dependence of the form exp( - jvt).
1.8 • Absorption of Light and Complex Refractive Index
61
Suppose that ko is the propagation constant in vacuum. This is a real quantity as a plane
wave suffers no loss in free space. The complex refractive index N with a real part n and imaginary part K is defined as the ratio of the complex propagation constant in a medium to propagation constant in free space.
that is,
N = n - jK = k>ko = (1>ko) 3k = - jk″4
(1.8.7)
Complex
refractive
index
n = k = >ko and K = k″>ko
The real part n is simply and generally called the refractive index and K is called the
­extinction coefficient. In the absence of attenuation
k″ = 0, k = k = and N = n = k>ko = k = >ko
We know that in the absence of loss, the relationship between the refractive index n and
the relative permittivity er is n = e1>2
r . This relationship is also valid in the presence of loss
­except that we must use a complex refractive index and complex relative permittivity, that is
N = n - jK = 1er = 1e′r - je″r
(1.8.8)
n2 - K 2 = er= and 2nK = e″r
(1.8.9)
By squaring both sides we can relate n and K directly to er= and e″r. The final result is
Optical properties of materials are typically reported either by showing the frequency
d­ ependences of n and K or er= and e″r. Clearly we can use Eq. (1.8.9) to obtain one set of properties
from the other.
Figure 1.26 shows the real (n) and imaginary (K ) parts of the complex refractive index of
CdTe as a function of wavelength in the infrared region to highlight their behavior around a resonance absorption phenomenon, when the energy transfer is maximum from the EM wave to the
material. Both ionic and electronic polarizations contribute to n (≈3.3) at low frequencies whereas
only electronic polarization contributes to n (≈2.6) at high frequencies. The extinction coefficient
peaks at about 72 om when the EM wave oscillations are efficiently coupled to the lattice vibrations,
Figure 1.26 Optical properties
of CdTe as a function of wavelength
in the infrared region.
Complex
refractive
index
Complex
refractive
index
62
Chapter 1 • Wave Nature of Light
that is, the vibrations of chains of Cd2+ and Te2- ions in the crystal,26 so that energy is passed from
the EM wave to these lattice vibrations. This type of absorption in which energy is passed to lattice
vibrations is called lattice or Reststrahlen absorption. Notice that K has a clear peak whereas
n shows a response that has a maximum, a minimum, and an inflection point (an S-like shape).
If we know the frequency dependence of the real part er= of the relative permittivity of a material, we can also determine the frequency dependence of the imaginary part e″r; and vice versa. This
may seem remarkable but it is true provided that we know the frequency dependence of either the
real or imaginary part over as wide a range of frequencies as possible (ideally from dc to infinity)
and the material is linear, that is, it has a relative permittivity that is independent of the applied
field; the polarization response must be linearly proportional to the applied field.27 The relationships that relate the real and imaginary parts of the relative permittivity are called Kramers-Kronig
relations, which involve integral transformations. If er= (v) and e″r(v) represent the frequency
dependences of the real and imaginary parts, then one can be determined from the other. Similarly,
if we know the wavelength dependence of n (or K), over as a wide wavelength range as possible,
we can determine the wavelength dependence of K (or n) using Kramers-Kronig relations for n-K.
It is instructive to mention that the reflection and transmission coefficients that we derived
above were based in using a real refractive index, that is neglecting losses. We can still use the
reflection and transmission coefficients if we simply use the complex refractive index N instead
of n. For example, consider a light wave traveling in free space incident on a material at normal
incidence (ui = 90°). The reflection coefficient is now
Reflection
coefficient
r =
1 - n + jK
1 + n - jK
(1.8.10)
The reflectance is then
R = `
Reflectance
n - jK - 1 2
(n - 1)2 + K 2
` =
n - jK + 1
(n + 1)2 + K 2
(1.8.11)
which reduce to the usual forms when the extinction coefficient K = 0.
The optical properties n and K can be determined by measuring the reflectance from the surface
of a material as a function of polarization and the angle of incidence (based on Fresnel’s equations).
Example 1.8.1
Complex refractive index of InP
An InP crystal has a refractive index (real part) n of 3.549 and an extinction coefficient K of 0.302 at a
wavelength of 620 nm (photon energy of 2 eV). Calculate the absorption coefficient a of InP at this wavelength and the reflectance of the air–InP crystal surface.
Solution
The absorption coefficient is
26
a = 2k″ = 2koK = 2 3(2p)>(620 * 10-9 m)4 (0.302) = 6.1 * 106 m-1
In physics, these would be optical phonons. The EM wave would be interacting with optical phonons, and passing its
energy onto these phonons.
27
In addition, the material system should be passive—contain no sources of energy.
1.9 • Temporal and Spatial Coherence
65
Figure 1.28 A Gaussian wave packet which has sinusoidal oscillations at a frequency yo and a Gaussian
envelope (amplitude variation) over time. Its coherence time ∆t between half maximum points is ∆t. Its frequency
spectrum is Gaussian, centered at the oscillation frequency yo, and extends over ∆y ≈ 1> ∆t.
way the oscillations were generated or emitted. For example, if one takes the envelope of the oscillations to be Gaussian as shown in Figure 1.28 so that the wave is a Gaussian wave packet, then the
Fourier transform will also be a Gaussian, centered at the oscillation frequency yo. The coherence
length and coherence time then usually refer to some suitable widths of the respective Gaussian
envelopes. If ∆y and ∆t are the full widths at half maximum (FWHM) spreads in the frequency and
time domains, as in Figure 1.28, then approximately29
∆y ≈
1
∆t
(1.9.2)
Equation (1.9.2) is usually known as the bandwidth theorem. Although a Gaussian wave packet
is useful in representing finite-length light wave packets, it still needs to be suitably truncated to
be able to represent more practical pulses since a Gaussian function extends over all times. This
truncation does not limit the use of Eq. (1.9.2) since it is meant to apply approximately to any wave
packet. Although the exact relationship in Eq. (1.9.2) depends on the shape of the wave packet,
Eq. (1.9.2) is still widely used for approximately relating ∆y and ∆t for various wave packets.
Coherence and spectral width are therefore intimately linked. For example, the orange
radiation at 589 nm (both D-lines together) emitted from a sodium lamp has spectral width
∆y ≈ 5 * 1011 Hz. This means that its coherence time is ∆t ≈ 2 * 10-12 s or 2 ps, and its
coherence length is 6 * 10-4 m or 0.60 mm. On the other hand, the red lasing emission from a
He-Ne laser operating in multimode has a spectral width around 1 * 109 Hz, which corresponds
to a coherence length of 30 cm. Furthermore, a continuous wave laser operating in a single mode
will have a very narrow linewidth and the emitted radiation will perhaps have a coherence length
of several hundred meters. Typically light waves from laser devices have substantial coherence
lengths and are therefore widely used in wave-interference studies and applications such as interferometry, holography, and laser Doppler anemometry. Suppose that standing at one location in
space we measure the field vs. time behavior shown in Figure 1.27 (c) in which the zero crossing
of the signal occurs randomly. Given a point P on this “waveform,” we cannot predict the “phase”
or the signal at any other point Q. Thus P and Q are not in any way correlated for any temporal separation except Q coinciding with P (or being very close to it by an infinitesimally short time interval).
There is no coherence in this “white” light signal and the signal essentially represents white noise;
29
FWHM is the width of the Gaussian function between the half maximum points (see Appendix A). The bandwidth
theorem for a Gaussian wave train is actually ∆y∆t = 0.88 which can be derived by using Fourier transforms.
Spectral
width and
coherence
time
1.10 • Superposition and Interference of Waves
Solution
The frequency and wavelength are related through y = c>l, so that differentiating the latter we can find
the frequency width ∆y from the wavelength width ∆l
∆y
dy
c
≈`
` = `- 2`
∆l
dl
l
so that
∆y = ∆l(c>l2) = (22 * 10-9 m)(3 * 108 m s-1)>(650 * 10-9 m)2
= 1.562 * 1013 Hz
Thus, the coherence time is
∆t ≈ 1> ∆y = 1>(1.562 * 1013 Hz) = 6.40 * 10-14 s or 64.0 fs
The coherence length is
lc = c∆t = 1.9 * 10-5 m or 19 microns
The above very short coherence length explains why LEDs are not used in interferometry.
1.10 Superposition and Interference of Waves
Optical interference involves the superposition of two or more electromagnetic waves in which
the electric field vectors are added; the fields add vectorially. The waves are assumed to be nearly
monochromic, and have to have the same frequency. Two waves can only interfere if they ­exhibit
mutual temporal coherence as in Figure 1.29 (a) at a point in space where they interact. Indeed,
­interference phenomena can be used to infer on the mutual coherence of the waves. When two
waves with the same frequency with fields E1 and E2 interfere, they generate a resultant field E
that corresponds to the superposition of individual fields, that is, E = E1 + E2. Consider
two linearly polarized plane waves that originate from O1 and O2, as schematically shown in
Figure 1.30, so that the field oscillations at some arbitrary point of interest P is given by
E1 = Eo1 sin (vt - kr1 - f 1) and E2 = Eo2 sin (vt - kr2 - f 2)
(1.10.1)
where r1 and r2 are the distances from O1 and O2 to P. These waves have the same v and k. Due
to the process that generates the waves, there is a constant phase difference between them given
Figure 1.30 Interference
of two mutually coherent
waves of the same frequency
originating from sources O1 and
O2. We examine the resultant
at P. The resultant field E
depends on the phase angle d
which depends on the optical
path difference k(r2 - r1).
67
68
Chapter 1 • Wave Nature of Light
by f 2 - f 1. The resultant field at P will be the sum of these two waves, that is, E = E1 + E2.
Its irradiance depends on the time average of E # E, that is, E # E, so that
E # E = (E1 + E2) # (E1 + E2) = E21 + E22 + 2E1 # E2
It is clear that the interference effect is in the 2E1 # E2 term. We can simply the above equation a
little further by assuming Eo1 and Eo2 are parallel with magnitudes Eo1 and Eo2. Further, irradiance
of the interfering waves are I1 = 12 ceoE 2o1 and I2 = 12 ceoE 2o2 so that the resultant irradiance is given
by the sum of individual irradiances, I1 and I2, and has an additional third term I21, that is,
Interference
Interference
for mutually
coherent
beams
Constructive
and
destructive
interference
Irradiance
of two
superimposed
incoherent
beams
I = I1 + I2 + 2(I1I2)1>2 cos d
(1.10.2)
where the last term is usually written as 2(I1I2)1>2 cos d = I21, and d is a phase difference given by
d = k(r2 - r1) + (f 2 - f 1)
(1.10.3)
Since we are using nearly monochromatic waves, (f 2 - f 1) is constant, and the interference therefore depends on the term k(r2 - r1), which represents the phase difference between
the two waves as a result of the optical path difference between the waves. As we move point P,
k(r2 - r1) will change because the optical path difference between the two waves will change;
and the interference will therefore also change.
Suppose (f 2 - f 1) = 0, the two waves are emitted from a spatially coherent source. Then,
if the path difference k(r2 - r1) is 0, 2p or a multiple of 2p, that is, 2mp, m = 0, {1, {2 c,
then the interference intensity I will be maximum; such interference is defined as constructive
interference. If the path difference k(r2 - r1) is p or 3p or an odd multiple of p, (2m + 1)p,
then the waves will be 180° out of phase, and the interference intensity will be minimum; such
interference is defined as destructive interference; both constructive and destructive intensity
are shown in Figure 1.30. The maximum and minimum irradiances are given by
Imax = I1 + I2 + 2(I1I2)1>2 and Imin = I1 + I2 - 2(I1I2)1>2
(1.10.4)
If the interfering beams have equal irradiances, then Imax = 4I1 and Imin = 0.
It is important to emphasize that we have considered the interference of two nearly monochromatic waves that exhibited mutual temporal and spatial coherence. If the waves do not
have any mutual coherence, that is, they are incoherent, then we cannot simply superimpose the
electric fields as we did above, and the detector at P, which averages the measurement over its
response time, will register an irradiance that is simply the sum of individual irradiances,
I = I1 + I2
(1.10.5)
One of the most described interference experiments is Young’s two slit experiment that
generates an interference fringe. In the modern version of this, a coherent beam of light, as
available from a laser, is incident on two parallel slits S1 and S2. There is a screen far away
from the slits on which the waves emanating from the slits interfere, as shown in Figure 1.31.
The result is an interference pattern that is composed of light and dark regions, corresponding to Imax and Imin. Since S1 and S2 are excited by the same wavefront, they emit coherent
waves, and we can take f 2 - f 1 = 0. Consider a point P at a distance y on the screen. The
phase difference d at P is then k(r2 - r1). As we move P along y, d = k(r2 - r1) changes
and the irradiance on the screen goes through minima and maxima with distance y, following
1.11 • Multiple Interference and Optical Resonators
69
Figure 1.31 Young’s two slit experiment. Slits S1 and S2, separated by s are illuminated at the same time
by coherent (nearly monochromatic) collimated laser beam. The irradiance at the screen shows bright and dark
fringes due to the interference of waves emanating from the two slits. The screen is assumed to be far away at a
distance L from the slits (L >> s).
Eq. (1.10.2), generating dark and light bands or fringes. The maxima occur when d = 2m p,
and minima when d = (2m + 1)p, where m = 0, {1, {2. It is not difficult to show that, if
the screen is far away, (r2 - r1) ≃ (s>L)y in which L is the distance from the slits to the screen,
and s is the separation of the slits, so that d is proportional to y. Assuming equal ­irradiances are
emitted from S1 and S2, I1 + I2 = Io, the brightness on the screen changes with y periodically as
I = Io 51 + cos 3(s>L)ky46
(1.10.6)
The resulting periodic interference pattern on the screen, as shown in Figure 1.31, is often
called Young’s interference fringes. In a better treatment one needs to consider not only the
­coherence length of the waves from S1 and S2, but also the diffraction that takes place at each slit
due to the finite width of the slit. (See diffraction in Section 1.12.)
1.11 Multiple Interference and Optical Resonators
Charles Fabry (1867–1945), left, and Alfred Perot (1863–1925), right, were the first French physicists to construct an optical cavity for interferometry. (Perot: The Astrophysical Journal, Vol. 64, November 1926, p. 208,
courtesy of the American Astronomical Society. Fabry: Courtesy of Library of Congress Prints and Photographs
Division, Washington, DC 20540, USA.)
Interference
72
Chapter 1 • Wave Nature of Light
intensity, as defined in Figure 1.32 (c). It can be calculated in a straightforward fashion when
R 7 0.6 from
Spectral
width and
Finesse
dym =
yf
F
; F =
pR1>2
1 - R
(1.11.5)
in which F is called the finesse of the resonator, which increases as losses decrease (R increases).
Large finesses lead to sharper mode peaks. Finesse is the ratio of mode separation (∆ym) to
spectral width (dym).
We can also define a quality factor Q for the optical resonant cavity in a similar fashion to
defining a Q-factor for an LC oscillator, that is,
Q-factor
and
Finesse
Quality factor, Q =
Resonant frequency
ym
=
= mF
Spectral width
dym
(1.11.6)
The Q-factor is a measure of the frequency selectiveness of a resonator; the higher the Q-factor,
the more selective the resonator, or narrower the spectral width. It is also a measure of the energy
stored in the resonator per unit energy dissipated (due to losses such as from the reflecting surfaces) per cycle of oscillation.
The Fabry–Perot optical cavities are widely used in laser, interference filter, and spectroscopic applications. Consider a light beam that is incident on a Fabry–Perot cavity as in
Figure 1.33. The optical cavity is formed by partially transmitting and reflecting plates. Part of
the incident beam enters the cavity. We know that only special cavity modes are allowed to exist
in the cavity since other wavelengths lead to destructive interference. Thus, if the incident beam
has a wavelength corresponding to one of the cavity modes, it can sustain oscillations in the cavity
and hence lead to a transmitted beam. The output light is a fraction of the light intensity in the
cavity and is proportional to Eq. (1.11.3). Commercial interference filters are based on this principle except that they typically use two cavities in series formed by dielectric mirrors (a stack
of quarter wavelength layers); the structure is more complicated than in Figure 1.33. Further,
adjusting the cavity length L provides a “tuning capability” to scan different wavelengths.
Equation (1.11.3) describes the intensity of the radiation in the cavity. The intensity of
the transmitted radiation in Figure 1.33 can be calculated, as above, by considering that each
time a wave is reflected at the right mirror, a portion of it is transmitted, and that these transmitted waves can interfere only constructively to constitute a transmitted beam when kL = mp.
Intuitively, if Iincident is the incident light intensity, then a fraction (1 - R) of this would enter
Figure 1.33
Transmitted light through a Fabry–Perot optical cavity.
1.11 • Multiple Interference and Optical Resonators
73
Left: Fused silica etalon. Right:
A 10 GHz air spaced etalon with
3 zerodur spacers. (Courtesy of
Light Machinery Inc.)
the cavity to build up into Icavity in Eq. (1.11.3), and a fraction (1 - R) of Icavity would leave the
cavity as the transmitted intensity Itansmitted. Thus,
Itransmitted = Iincident
(1 - R)2
(1 - R)2 + 4R sin2 (kL)
(1.11.7)
Transmitted
mode
intensities
which is again maximum just as for Icavity whenever kL = mp as shown in terms of wavelength
in Figure 1.33.31
The ideas above can be readily extended to a medium with a refractive index n by using nk
for k or l>n for l where k and l are the free-space propagation constant and wavelength, respectively. Equations (1.11.1) and (1.11.2) become
ma
l
b = L m = 1, 2, 3, c
2n
ym = m a
c
b = myf ; yf = c>(2nL)
2nL
(1.11.8)
Fabry–Perot
cavity modes
in a medium
(1.11.9)
Cavity
resonant
frequencies
Further, if the angle of incidence u at the etalon face is not normal, then we can resolve k to
be along the cavity axis; that is use k cos u instead of k in the discussions above.
The two mirrors in Figure 1.32 (a) were assumed to have the same reflectance R. Suppose
that R1 and R2 are the reflectances of the mirrors M1 and M2. Then, we can continue to use the
above equations by using an average geometric reflectance, that is R = (R1R2)1>2.
Example 1.11.1 Resonator modes and spectral width
of a semiconductor Fabry–Perot cavity
Consider a Fabry–Perot optical cavity made of a semiconductor material with mirrors at its ends. (The mirrors have been obtained by coating the end of the semiconductor crystal.) The length of the semiconductor,
and hence the cavity, is 250 om and mirrors at the ends have a reflectance of 0.90. Calculate the cavity
mode nearest to the free-space wavelength of 1310 nm. Calculate the separation of the modes, finesse,
spectral width of each mode in frequency and wavelength, and the Q-factor.
31
The term on the right multiplying Iincident in Eq. (1.11.7) is usually known as the Airy function.
76
Chapter 1 • Wave Nature of Light
Figure 1.35 (a) Huygens-Fresnel principle states that each point in the aperture becomes a source of
secondary waves (spherical waves). The spherical wavefronts are separated by l. The new wavefront is the
envelope of all these spherical wavefronts. (b) Another possible wavefront occurs at an angle u to the z-direction,
which is a diffracted wave.
waves arriving from all point sources in the aperture. The screen is far away from the aperture
so the waves arrive almost parallel at the screen (alternatively a lens can be used to focus the
diffracted parallel rays to form the diffraction pattern).
Consider an arbitrary direction u, and consider the phase of the emitted wave (Y) from an
arbitrary point source at y with respect to the wave (A) emitted from source at y = 0 as shown
in Figure 1.36 (a). If k is the propagation constant, k = 2p>l, the wave Y is out of phase with
respect to A by ky sin u. Thus the wave emitted from the point source at y has a field dE,
dE ∝ (dy) exp (-jky sin u)
(1.12.1)
All of these waves from point sources from y = 0 to y = a interfere at the screen at a
point P that makes an angle u at the slit, and the resultant field at the screen at this point P is
Figure 1.36 (a) The aperture has a finite width a along y, but it is very long along x so that it is a onedimensional slit. The aperture is divided into N number of point sources each occupying dy with amplitude
proportional to dy since the slit is excited by a plane electromagnetic wave. (b) The intensity distribution in the
received light at the screen far away from the aperture: the diffraction pattern. Note that the slit is very long
along x so that there is no diffraction along this dimension. The incident wave illuminates the whole slit.
(c) Typical diffraction pattern using a laser pointer on a single slit. The difference from the pattern in (b) is due to
the finite size of the laser pointer beam along x that is smaller than the length of the slit.
1.12 • Diffraction Principles
77
their sum. Because the screen is far away, a point on the screen is at the same distance from
anywhere in the aperture. This means that all the spherical waves from the aperture experience
the same phase change and decrease in amplitude in reaching the screen. This simply scales dE
at the screen by an amount that is the same for all waves coming from the aperture. Thus, the
resultant field E(u) at point P at the screen is
y=a
E(u) = C
Ly = 0
dy exp ( -jky sin u)
(1.12.2)
in which C is a constant. Integrating Eq. (1.12.2) we get
Ce-j2 ka sin u a sin 112 ka sin u2
1
E(u) =
1
2
ka sin u
The light intensity I at a point at P at the screen is proportional to Eu 2, and thus
I(u) = £
C = a sin 112 ka sin u2
1
2
ka sin u
2
§ = I(0) sinc2 (b); b = 12 (ka sin u)
(1.12.3)
Single slit
diffraction
equation
in which C = is a constant and b is a convenient new variable representing u, and sinc (“sink”) is
a function that is defined by sinc (b) = sin (b)>(b).
If we were to plot Eq. (1.12.3) as a function of u at the screen we would see the intensity (diffraction) pattern schematically depicted in Figure 1.36 (b). First, observe that the pattern has bright
and dark regions, corresponding to constructive and destructive interference of waves emanating
from the aperture. Second, the center bright region is wider than the aperture width a, which mean
that the transmitted beam must be diverging. The zero intensity occurs when, from Eq. (1.12.3),
sin u =
ml
; m = {1, {2, c
a
(1.12.4)
The angle uo for the first zero, corresponding to m = {1, is given by uo = {l>a, where
we assumed that the divergence is small (usually the case) so that sin uo ≈ uo. Thus, the divergence ∆u, the angular spread, of the diffracted beam is given by
∆u = 2uo ≈
2l
a
(1.12.5)
A light wave at a wavelength 1300 nm, diffracted by a slit of width a = 100 om (about
the thickness of this page), has a divergence ∆u of about 1.5°. From Figure 1.36 (b), it is apparent that, using geometry, we can easily calculate the width c of the central bright region of the
intensity pattern, given uo from Eq. (1.12.5) and the distance R of the screen from the aperture.
The diffraction patterns from two-dimensional apertures such as rectangular and circular
apertures are more complicated to calculate but they use the same principle based on the multiple interference of waves emitted from all point sources in the aperture. The diffraction pattern
of a rectangular aperture is shown in Figure 1.37. It involves the multiplication of two individual single slit (sinc) functions, one slit of width a along the horizontal axis, and the other of
width b along the vertical axis. (Why is the diffraction pattern wider along the horizontal axis?)
The diffraction pattern from a circular aperture, known as Airy rings, was shown in
Figure 1.34, and can be roughly visualized by rotating the intensity pattern in Figure 1.36 (b)
Zero
intensity
points
Divergence
from single
slit of
width a
78
Chapter 1 • Wave Nature of Light
Figure 1.37 The rectangular
aperture of dimensions a * b
on the left gives the diffraction
pattern on the right (b is twice a).
about the z-axis. We can, as we did for the single slit, sum all waves emanating from every point
in the circular aperture, taking into account their relative phases when they arrive at the screen
to obtain the actual intensity pattern at the screen. The result is that the diffraction pattern is a
Bessel function of the first kind,36 and not a simply rotated sinc function. The central white spot
is called the Airy disk; its radius corresponds to the radius of the first dark ring. We can still
use Figures 1.36 (a) and (b) to imagine how diffraction occurs from a circular aperture by taking
this as a cut through the aperture so that a is now the diameter of the aperture, denoted as D. The
angular position uo of the first dark ring, as defined as in Figure 1.36 (b), is determined by the
diameter D of the aperture and the wavelength l, and is given by
Angular
radius of
Airy disk
sin uo = 1.22
l
D
(1.12.6)
The divergence angle from the aperture center to the Airy disk circumference is 2uo. If R is
the distance of the screen from the aperture, then the radius of the Airy disk, approximately b, can
be calculated from the geometry in Figure 1.36 (b), which gives b>R = tan uo ≈ uo. If a lens is
used to focus the diffracted light waves onto a screen, then R = f, focal length of the lens.
It is worth commenting on the Gaussian beam at this point. Suppose we now examine a Gaussian beam with a waist 2wo that is the same as the aperture size D. The far field
Diffraction pattern far away from a
circular aperture.
Diffraction pattern far away from a square
aperture.
36
Bessel functions are special mathematical functions, which can be looked up in mathematics handbooks. They are used
in various engineering problems.
1.12 • Diffraction Principles
79
half-­divergence angle u of this Gaussian beam would be u = (2/p)(l>D) or 0.637(l>D) as in
Eq. (1.1.7). The Gaussian beam has a smaller divergence than the diffracted beam from a circular aperture. The difference is due the fact that each point in the circular aperture emits with
the same intensity because the aperture is illuminated by a plane wave. If we were to change
the emission intensity within the aperture to follow a Gaussian distribution, we would see a
Gaussian beam as the “diffracted beam.” The Gaussian beam is a self-diffracted beam and has
the smallest divergence for a given beam diameter.
Example 1.12.1 Resolving power of imaging systems
Consider what happens when two neighboring point light sources are examined through an ­imaging system
with an aperture of diameter D (this may even be a lens). The two sources have an angular separation of ∆u
at the aperture. The aperture produces a diffraction pattern of the sources S1 and S2, as shown in Figure 1.38.
As the points get closer, their angular separation ­becomes narrower and the diffraction patterns overlap more.
According to the Rayleigh criterion, the two spots are just resolvable when the principal maximum of one
diffraction pattern coincides with the minimum of the other, which is given by the condition
sin (∆u min ) = 1.22
l
D
(1.12.7)
The human eye has a pupil diameter of about 2 mm. What would be the minimum angular separation
of two points under a green light of 550 nm and their minimum separation if the two objects are 30 cm from
the eye? The image will be a diffraction pattern in the eye, and is a result of waves in this medium. If the
refractive index n ≈ 1.33 (water) in the eye, then Eq. (1.12.7) is
sin (∆umin) = 1.22
(550 * 10-9 m)
l
= 1.22
nD
(1.33)(2 * 10-3 m)
giving
∆umin = 0.0145°
Their minimum separation s would be
s = 2L tan (∆umin >2) = 2(300 mm) tan (0.0145°>2) = 0.076 mm = 76 om
which is about the thickness of a human hair (or this page).
Figure 1.38 Resolution of imaging systems is limited by diffraction effects. As points S1 and S2 get closer,
eventually the Airy patterns overlap so much that the resolution is lost. The Rayleigh criterion allows the
minimum angular separation of two of the point sources to be determined. (Schematic illustration inasmuch as
the side lobes are actually much smaller than the center peak.)
Angular
limit of
resolution
80
Chapter 1 • Wave Nature of Light
Image of two-point sources captured through a small circular aperture. (a) The two points are fully resolved since the diffraction patterns of the two sources are sufficiently separated. (b) The two images are near the Rayleigh limit of resolution.
(c) The first dark ring of one image passes through the center of the bright Airy disk of the other. (Approximate.)
B. Diffraction Grating
A diffraction grating in its simplest form is an optical device that has a periodic series of slits in an
opaque screen as shown in Figure 1.39 (a). An incident beam of light is diffracted in certain welldefined directions that depend on the wavelength l and the grating properties. Figure 1.39 (b) shows a
typical intensity pattern in the diffracted beam for a finite number of slits. There are “strong beams
of diffracted light” along certain directions (u) and these are labeled according to their occurrence:
zero-order (center), first-order, either side of the zero-order, and so on. If there are an infinite number of slits then the diffracted beams have the same intensity. In reality, any periodic variation in
the refractive index would serve as a diffraction grating and we will discuss other types later. As in
Fraunhofer diffraction we will assume that the observation screen is far away, or that a lens is used
to focus the diffracted parallel rays on to the screen (the lens in the observer’s eye does it naturally).
We will assume that the incident beam is a plane wave so that the slits become coherent
(synchronous) sources. Suppose that the width a of each slit is much smaller than the separation
Figure 1.39 (a) A diffraction grating with N slits in an opaque screen. Slit periodicity is d and slit width is a;
a V d. (b) The diffracted light pattern. There are distinct, that is diffracted, beams in certain directions (schematic).
(c) Diffraction pattern obtained by shining a beam from a red laser pointer onto a diffraction grating. The finite size
of the laser beam results in the dot pattern. (The wavelength was 670 nm, red, and the grating has 200 lines per inch.)
1.12 • Diffraction Principles
81
d of the slits as shown in Figure 1.39 (a). Waves emanating at an angle u from two neighboring slits are out of phase by an amount that corresponds to an optical path difference d sin u.
Obviously, all such waves from pairs of slits will interfere constructively when this is a multiple
of the whole wavelength
d sin u = ml; m = 0, {1, {2, c
(1.12.8)
which is the well-know grating equation, also known as the Bragg37 diffraction condition.
The value of m defines the diffraction order; m = 0 being zero-order, m = {1 being first-order,
etc. If the grating in Figure 1.39 (a) is in a medium of refractive index n, that is, the incident and
diffracted beams are all in the same medium of index n, then we should use l>n for the wavelength in Eq. (1.12.8), where l is the free-space wavelength, that is, d sin u = ml>n.
The problem of determining the actual intensity of the diffracted beam is more complicated as it involves summing all such waves at the observer and, at the same time, including the
diffraction effect of each individual narrow slit. With a smaller than d as in the Figure 1.39 (a),
the amplitude of the diffracted beam is modulated by the diffraction amplitude of a single slit
since the latter is spread substantially, as illustrated in Figure 1.39 (b). It is apparent that the diffraction grating provides a means of deflecting an incoming light by an amount that depends on
its wavelength—the reason for their use in spectroscopy.
The diffraction grating in Figure 1.40 (a) is a transmission grating. The incident and
diffracted beams are on opposite sides of the grating. Typically, parallel thin grooves on a glass
plate would serve as a transmission grating as in Figure 1.40 (a). A reflection grating has the
incident beam and the diffracted beams on the same side of the device as in Figure 1.40 (b).
The surface of the device has a periodic reflecting structure, easily formed by etching parallel
grooves in a metal film, etc. The reflecting unetched surfaces serve as synchronous secondary
sources that interfere along certain directions to give diffracted beams of zero-order, first-order,
etc. Among transmission gratings, it is customary to distinguish between amplitude gratings in
which the transmission amplitude is modulated, and so-called phase gratings where only the
refractive index is modulated, without any losses.
Figure 1.40 (a) Ruled periodic parallel scratches on a glass serve as a transmission grating. (The glass plate is
assumed to be very thin.) (b) A reflection grating. An incident light beam results in various “diffracted” beams. The
zero-order diffracted beam is the normal reflected beam with an angle of reflection equal to the angle of incidence.
37
William Lawrence Bragg (1890–1971), Australian-born British physicist, won the Nobel Prize with his father, William
Henry Bragg, for his “famous equation” when he was only 25 years old.
Grating
equation
82
Chapter 1 • Wave Nature of Light
When the incident beam is not normal to the diffraction grating, then Eq. (1.12.8) must be
modified. If ui is the angle of incidence with respect to the normal to the grating, then the diffraction angle um for the m-th mode is given by
Grating
equation
d(sin um - sin ui) = ml; m = 0, {1, {2, c
(1.12.9)
The same equation can be used for transmission and reflection gratings provided that we
define the angles ui and um as positive on either side of the normal as in Figure 1.40 (b).38
Consider a grating with N slits. The slit width is a (very narrow), and d is the periodicity as
before. The detector is at a distance L, far away from the grating. While the periodicity in the
slits gives rise to the diffracted beams, the diffraction at each narrow slit defines the envelope of
the diffracted intensities as shown in Figure 1.39 (b). If the incident plane wave is normal to the
grating, the intensity distribution along y at the screen is given by
Grating
diffraction
pattern
I(y) = Io £
sin 112 ky a2
1
2 kya
2
§ £
sin 1 12 Nkyd2
Nsin 1 12 kyd 2
2
§
(1.12.10)
where ky is the scattering wave vector defined by ky = (2p>l)(y>L) = (2p>l)sin u, and Io is the
maximum intensity along u = 0. The ­second term represents the oscillations in the intensity due
to interference from different slits. The first term is the envelope of the diffraction pattern, and is
the diffraction pattern of a single slit.
The resolvance or the resolving power R of a diffraction grating is its ability to be able
to separate out adjacent wavelengths. If l2 - l1 = ∆l is the minimum wavelength separation
that can be measured, as determined by the Rayleigh criterion (the maximum of the intensity
distribution at l1 is at the first minimum of the intensity distribution at l2), and l is the average
wavelength (1>2)(l1 + l2) in ∆l, then the resolving power is defined by
Resolving
power
Blazing
angle
R = l> ∆l
(1.12.11)
The separation ∆l is also called the spectral resolution. If N grooves on a grating are
i­lluminated and the order of diffraction is m, the theoretical resolving power is given simply by
R = mN. The resolving power is also called the chromatic resolving power since it refers to
the separation of wavelengths.
Diffraction gratings are widely used in spectroscopic applications because of their ability
to provide light deflection that depends on the wavelength. In such applications, the undiffracted
light that corresponds to the zero-order beam (Figure 1.40) is clearly not desirable because it
wastes a portion of the incoming light intensity. Is it possible to shift this energy to a higher
order? Robert William Wood (1910) was able to do so by ruling grooves on glass with a controlled shape as in Figure 1.41 (a) where the surface is angled periodically with a spatial period d.
The diffraction condition in Eq. (1.12.9) applies with respect to the normal to the grating plane,
whereas the first-order reflection corresponds to reflection from the flat surface, which is at an
angle g. Thus it is possible to “blaze” one of the higher orders (usually m = 1) by appropriately choosing g. Most modern diffraction gratings are of this type. If the angle of incidence is
ui with respect to the grating normal, then specular reflection occurs at an angle (g + ui) with
respect to the face normal and (g + ui) + g with respect to the grating normal. This reflection at
(g + ui) + g should occur at diffraction angle um so that
2g = um - ui
(1.12.12)
38
Some books use d(sin um + sin ui) = ml for a transmission grating but the angles become positive on the incidence
side and negative on the transmitted side with respect to the normal. It is a matter of sign convention.
1.14 • Thin Film Optics: Multiple Reflections in Thin Films
87
The amplitude of the reflected beam is
Areflected = A1 + A2 + A3 + A4 + c
that is,
Areflected >A0 = r1 + t1t 1= r2e-jf - t1t 1= r1r 22e-j2f + t1t 1= r 21r 32e-j3f + c(1.14.2)
which is a geometric series. Using Eq. (1.14.1c), Eq. (1.14.2) can be conveniently summed to
obtain the overall reflection coefficient r
r =
r1 + r2e-jf
1 + r1r2e
-jf
(1.14.3)
Thin film
reflection
coefficient
Similarly, we can sum for the amplitude of the transmitted beam as
Ctransmitted = C1 + C2 + C3 + c
that is,
Ctransmitted >A0 = t1t2e-jf>2 - t1t2r1r2e-j3f>2 + t1t2r 21r 22e-j5f>2 + c
(1.14.4)
which is a geometric series that sums to
t =
t1t2e-jf>2
1 + r1r2e
-jf
(1.14.5)
Equations (1.14.3) and (1.14.5) describe the reflected and transmitted waves. The reflectance and transmittance are then
R = r 2 T = (n3 >n1) t 2
(1.14.6)
l
(2m + 1); m = 0, 1, 2, c
4n2
(1.14.7)
Figure 1.46 (a) shows R and T as a function of f for n1 6 n2 6 n3. Clearly, R is minimum,
and T is maximum, whenever f = p * (odd number) or f = 2(2p>l)n2d = p(2m + 1),
where m = 0, 1, 2, c The latter leads to
d =
which is the thickness required to minimize the reflection, and maximize the transmission of
light when n2 is intermediate between n1 and n3. The oscillations in R and T with f (e.g., as the
wavelength is scanned) are sometime referred to as an interference fringes in wavelength.
Figure 1.46 (a) Reflectance R and transmittance T vs. f = 2n2d>l, for a thin film on a substrate where
n1 = 1 (air), n2 = 2.5, n3 = 3.5, and n1 6 n2 6 n3. (b) R and T vs. f for a thin film on a substrate where
n1 = 1 (air), n2 = 3.5, n3 = 2.5, and n2 7 n3 7 n1.
Thin film
transmission
coefficient
Thin film
reflectance
and
trans­mittance
Thickness
for minimum
reflection
88
Chapter 1 • Wave Nature of Light
Figure 1.46 (b) represents the reflectance and transmittance vs. f of light through a thin layer
of high index material on a low index substrate where n1 6 n3 6 n2; for example, a semiconductor
film on a glass substrate. Notice the difference between the two cases, especially the locations of the
maxima and minima. (Why is there a difference?)
The phase change f, of course, depends on three factors, d, n2, and l. The minimum
and maximum reflectances in Figure 1.46 at those particular f values can be found by using
Eqs. (1.14.3) and (1.14.7). For n1 6 n2 6 n3 as in Figure 1.46 (a)
Minimum
and
maximum
reflectance
Rmin = a
n22 - n1n3
n22 + n1n3
2
b ; Rmax = a
n3 - n1 2
b
n3 + n1
(1.14.8)
and the transmittance can be found from R + T = 1. When n1 6 n3 6 n2 then Rmin and Rmax
equations are interchanged. While Rmax appears to be independent from n2, the index n2 is nonetheless still involved in determining maximum reflection inasmuch as R reaches Rmax when
f = 2(2p>l)n2d = p(2m); when f = p * (even number).
In transmission spectra measurements, a spectrophotometer is used to record the transmittance of a light beam as a function of wavelength through a sample. If the sample is a thin film
on a substrate, there will be multiple reflections and interferences in the thin film, and the measured transmittance will exhibit maxima and minima as in Figure 1.46 as the wavelength (or f)
is scanned. The locations of the maxima and minima, with the knowledge of the substrate index
(n3), can be used to find d and n.
Example 1.14.1 Thin film optics
Consider a semiconductor device with n3 = 3.5 that has been coated with a transparent optical film
(a ­dielectric film) with n2 = 2.5, n1 = 1 (air). If the film thickness is 160 nm, find the minimum and maximum reflectances and transmittances and their corresponding wavelengths in the visible range. (Assume
normal incidence.)
Solution
This case corresponds to Figure 1.46 (a). Minimum reflectance Rmin occurs at f = p or odd multiple of p,
and maximum reflectance Rmax at f = 2p or an integer multiple of 2p. From Eq. (1.14.8) we have
and
Rmin = a
Rmax = a
n22 - n1n3
n22
+ n1n3
2
b = a
2.52 - (1)(3.5)
2
2.5 + (1)(3.5)
2
b = 0.080 or 8.0,
n3 - n1 2
3.5 - 1 2
b = a
b = 0.31 or 31,
n3 + n1
3.5 + 1
Corresponding transmittances are,
Tmax = 1 - R min = 0.92 or 92,
and
Tmin = 1 - R max = 0.69 or 69,
Without the thin film coating, the reflectance would be 31%, the maximum reflectance.
90
Transmittance
for a thick
plate for
incoherent
light
Transmittance
for a thick
plate or
incoherent
beam of
light
Chapter 1 • Wave Nature of Light
By substituting for R in terms of the indices, we can write this as
Tplate =
2n1n2
n21
+ n22
(1.15.2)
For example, for a glass plate of n2 = 1.60 in air (n1 = 1), Tplate = 89.9, while the simple transmittance through an n1 9n2 interface would give 94.7%. The overall reflectance is 1 - Tplate so that
Rplate =
(n1 - n2)2
n21 + n22
(1.15.3)
One of the simplest ways to determine the refractive index of a plate is to measure the transmittance Tplate in Eq. (1.15.1), from which we can calculate n2.
1.16 Scattering of Light
When a light beam propagates in a medium in which there are small particles or inhomogeneities, such as local changes in the refractive index of the medium, some of the power in the beam
is radiated away from the direction of propagation, that is some of the power becomes scattered.
Scattering is a process by which some of the power in a propagating electromagnetic wave is
redirected as secondary EM waves in various directions away from the original direction of
propagation as illustrated in Figure 1.48 (a). There are a number of scattering processes, which
are usually classified in terms of the size of the scattering particles in relation to the wavelength
of light that is scattered. In Rayleigh scattering, the scattering particle size, or the scale of
­inhomogeneities in a medium, is much smaller than the wavelength of light. The intensity of the
scattered light at an angle u to the original beam depends on the scattering process; the Rayleigh
scattering case is shown in Figure 1.48 (b) in which the scattering is not spherically symmetric.
Consider what happens when a propagating wave encounters a molecule, an impurity
in a crystal or a small dielectric particle (or region), which is smaller than the wavelength of
light. The electric field in the wave polarizes the particle by displacing the lighter electrons with
­respect to the heavier positive nuclei. The electrons in the molecule couple and oscillate with
the electric field in the wave (ac electronic polarization). The oscillation of charge “up” and
“down,” or the oscillation of the induced dipole, radiates EM waves all around the molecule as
illustrated in Figure 1.48 (a). We should remember that an oscillating charge is like an alternating current which always radiates EM waves (like an antenna). The net effect is that the incident
Figure 1.48 (a) Rayleigh scattering involves the polarization of a small dielectric particle or a region that is
much smaller than the light wavelength. The field forces dipole oscillations in the particle (by polarizing it) which
leads to the emission of EM waves in “many” directions so that a portion of the light energy is directed away from the
incident beam. (b) A polar plot of the dependence of the intensity of the scattered light on the angular direction u with
respect to the direction of propagation x in Rayleigh scattering (in a polar plot, the radial distance OP is the intensity).
1.16 • Scattering of Light
Lord Rayleigh (John William Strutt) was an English physicist
(1877–1919) and a Nobel Laureate (1904) who made a number of
contributions to wave physics of sound and optics. He formulated
the theory of scattering of light by small particles and the dependence of scattering on 1/l4 circa 1871. Then, in a paper in 1899
he provided a clear explanation on why the sky is blue. Ludvig
Lorentz, around the same time, and independently, also formulated
the scattering of waves from a small dielectric particle, though
it was published in Danish (1890).40 (© Mary Evans Picture
Library/Alamy.)
wave becomes partially reradiated in different directions and hence loses intensity in its original
direction of propagation. (We may think of the process as the particle absorbing some of the
energy via electronic polarization and reradiating it in different directions.) It may be thought
that the scattered waves constitute a spherical wave emanating from the scattering molecule, but
this is not generally the case as the re-emitted radiation depends on the shape and polarizability
of the molecule in different directions. We assumed a small particle so that at any time the field
has no spatial variation through the particle, whose polarization then oscillates with the electric
field ­oscillation. Whenever the size of the scattering region, whether an inhomogeneity, a small
­particle, a molecule, or a defect in a crystal, is much smaller than the wavelength l of the incident wave, the scattering process is generally termed Rayleigh scattering. Typically, the particle
size is smaller than one-tenth of the light wavelength.
Rayleigh scattering of light propagating in a glass medium is of particular interest
in photonics because it results in the attenuation of the transmitted light pulses in optical
­fibers. The glass structure is such that there are small random spatial variations in the refractive index about some average value. There are therefore local fluctuations in the relative
­permittivity and polarizability, which effectively act if there are small inhomogeneities in the
medium. These dielectric inhomogeneities arise from fluctuations in the relative permittivity
that is part of the intrinsic glass structure. As the fiber is drawn by freezing a liquid-like flow,
random thermodynamic fluctuations in the composition and structure that occur in the liquid
state become frozen into the solid structure. Consequently, the glass fiber has small fluctuations in the relative permittivity which leads to Rayleigh scattering. A small inhomogeneous
region acts like a small dielectric particle and scatters the propagating wave in different
­directions. Nothing can be done to eliminate Rayleigh scattering in glasses as it is part of
their intrinsic structure.
It is apparent that the scattering process involves electronic polarization of the molecule
or the dielectric particle. We know that this process couples most of the energy at ultraviolet
frequencies where the dielectric loss due to electronic polarization is maximum and the loss is
due to EM wave radiation. Therefore, as the frequency of light increases, the scattering becomes
more severe. In other words, scattering decreases with increasing wavelength. The intensity
of the scattered radiation is proportional to 1>l4. For example, blue light, which has a shorter
wavelength than red light, is scattered more strongly by air molecules. When we look at the sun
directly, it appears yellow because the blue light has been scattered in the direct light more than
40
Pedro Lilienfeld’s “A Blue Sky History”, in Optics and Photonics News, 15(6), 32, 2004, is highly recommended; it
also provides the original references.
91
1.17 • Photonic Crystals
a general class of optical materials called photonic crystals. A photonic crystal (PC) is a material that has been structured to possess a periodic modulation of the refractive index n, just like
in Figure 1.24, so that the structure influences the propagation and confinement of light within it.
The periodicity can be in one- (1D), two- (2D), or three-dimensional (3D); Figures 1.50 (a)–(c)
illustrates 1D, 2D, and 3D simple photonic crystals as examples. In fact, quite complicated structures can be constructed that have very interesting optical properties. The dielectric mirror or
the Bragg reflector can be viewed as one of the simplest 1D photonic crystals.41 For the structure
to influence the propagation of the EM wave, it has to diffract the wave, which means that the
scale of periodic variations must be on the wavelength scale. As a­ pparent from Figure 1.24, the
1D PC has a band of frequencies over which it reflects the light and, conversely, there is a stop
band over which no transmission is possible through the dielectric stack. There is a band of
frequencies that represent waves that are not allowed to go through this periodic structure in the
direction of refractive index variation, along z in Figure 1.50; this band is called an optical or
photonic bandgap.
The periodic variation in n in Figure 1.50 is normally assumed to extend indefinitely,
whereas in practice, the PCs have a finite size, for example, a certain number of layers in the
dielectric mirror, not infinite. As in normal crystals, the periodic structures in Figure 1.50 have a
unit cell, which ­repeats itself to generate the whole lattice—that is, the whole crystal structure.
For the 1D PC in Figure 1.50 (a), for example, two adjacent layers, n1n2, form the unit cell. We
can move this unit cell along z by a distance Λ, the period (or periodicity), many times to generate the whole 1D photonic crystal.
The periodicity of a photonic crystal implies that any property at a location z will be the
same at z { Λ, z { 2Λ and so on; that is, there is translational symmetry along z (in 1D).
Figure 1.50 Photonic crystals in (a) 1D, (b) 2D, and (c) 3D, D being the dimension. Gray and white regions
have different refractive indices and may not necessarily be the same size. Λ is the periodicity. The 1D photonic
crystal in (a) is the well-known Bragg reflector, a dielectric stack.
41
The propagation of light through such a one-dimensional (1D) periodic variation of n has been well-known, dating
back to the early work of Lord Rayleigh in 1887. Eli Yablonovitch has suggested that the name “photonic crystal” should
­really only apply to 2D and 3D periodic structures with a large dielectric (refractive index) difference. (E. Yablonovitch,
“Photonic crystals: What’s in a name?,” Opt. Photon. News, 18, 12, 2007.) Nonetheless, the dielectric mirror in
Figure 1.50 (a) is often considered as the simplest 1D photonic crystal to derive the concept of a “photonic bandgap,”
essentially a stop band.
93
94
Chapter 1 • Wave Nature of Light
The EM waves that are allowed to propagate along z through the periodic structure are called the
modes of the photonic crystal. They have a special waveform that must bear the periodicity of
the structure, and are called Bloch waves. Such a wave for the field Ex, for example, has the form
Ex(z, t) = A(z) exp (-jkz), which represents a traveling wave along z and A(z) is an amplitude
function that has the periodicity of the structure, that is, it is periodic along z with a period Λ. A(z)
depends on the periodic refractive index function n(z). As we will see in Chapter 3, the electron
motion in a semiconductor crystal is also described by Bloch waves (electron wavefunctions).
As in the case of the dielectric mirror, the 1D PC has a band of frequencies over which
there can be no propagation along z. In a homogeneous medium of refractive index n, the relationship between the frequency v and the propagation constant k is simple, that is, c>n = v>k,
where k is the propagation constant inside the medium The dispersion behavior of the medium,
that is, v vs. k, is a straight line with a slope c>n. The dispersion characteristic of a 1D PC
for waves along z in Figure 1.50 (a) is shown in Figure 1.51 (a). We notice several important
characteristics. At low frequencies (long wavelengths), the waves propagate as if they are in a
homogenous medium with a constant phase velocity (dashed straight line). As expected, there
is a band of frequencies ∆v = v2 - v1 (between S1 and S2) over which no propagation along
z is allowed, which is a photonic bandgap along z. The whole v vs. k curves are periodic in
k with a period 2p>Λ. The point P is equivalent to P′ because k = k′ + (2p>Λ). We only need
to consider k-vales from -p>L to p>L. This region is called the first Brillouin zone.
At low frequencies or long wavelengths (small k values) in Figure 1.51 (a), the wavelength
is so much longer than the variations in n that the propagating wave experiences essentially some
average refractive index, nav, that is (n1 + n2)>2 if the n1 and n2 layers have the same thickness,
and propagates through the structure as if the structure was a homogenous medium with some
­effective refractive index, nav. Its phase velocity is v>k, which is c>nav, and its group velocity,
the slope of the v vs. k characteristic, is the same as c>nav, in this region (ignoring the wavelength dependence of n1 and n2). As the wavelength decreases (v increases), partial reflections of
the waves from the boundaries become important and interfere with propagation. At sufficiently
small wavelengths, diffraction becomes important as all these partially reflected waves interfere
with each other and give rise to significant reflection. Eventually, a critical wavelength (corresponding to v1) is reached where the waves become fully diffracted or reflected backwards.
A backward traveling (in -z direction) wave experiences the same reflection. These forward and
backward diffracted waves give rise to a standing wave in the structure; indeed, only the latter
can exist and waves cannot propagate, that is, travel freely.
Eli Yablonovitch at the University of California at Berkeley, and Sajeev John
(shown later in this chapter) at the University of Toronto, carried out the initial
pioneering work on photonic crystals. Eli Yablonovitch has suggested that the
name “photonic crystal” should apply to 2D and 3D periodic structures with
a large dielectric (refractive index) difference. (E. Yablonovitch, “Photonic
crystals: What’s in a name?,” Opt. Photon. News, 18, 12, 2007.) Their original
­papers were published in the same volume of Physical Review Letters in 1987.
According to Eli Yablonovitch, “Photonic Crystals are semiconductors for
light.” (Courtesy of Eli Yablonovitch.)
1.17 • Photonic Crystals
Figure 1.51 (a) Dispersion relation, v vs. k, for waves in a 1D PC along the z-axis. There are allowed modes and forbidden
modes. Forbidden modes occur in a band of frequencies called a photonic bandgap. (b) The 1D photonic crystal corresponding
to (a), and the corresponding points S1 and S2 with their stationary wave profiles at v1 and v2.
Consider what happens when reflections such as C and D from two successive unit cells
interfere constructively and give rise to a backward diffracted (reflected) wave as illustrated in
Figure 1.51 (b). Reflections from two successive unit cells must be in phase for full reflection.
This means that the phase difference between C and D (2p>l)(2n1d1 + 2n2d2) must be 2mp,
where m = 1, 2, c is an integer. If we define nav = (n1d1 + n2d2)>Λ, and k = nav(2p>l)
then reflection occurs when 2kΛ = 2mp, that is, k = mp>Λ. These are the wave vectors or
propagation constants of the waves that cannot be propagated. The waves suffer Bragg reflection
in 1D. What happens to these diffracted waves?
The diffracted waves in +z and -z directions set up a standing wave in the structure. If we
write the two reflected waves in opposite directions as A exp ( jkz) and A exp (-jkz) they would add as
A exp (jkz) { A exp (-jkz), which shows that there two possibilities S1 and S2. One of them, S1, has
most of its energy inside the n2, high refractive index layers, and hence has a lower frequency v1. S2
has most of its energy in the n1 layers and has a higher frequency v2; these are shown in Figure 1.51
(b). There are no waves in the v2 - v1 interval, which we know as the photonic bandgap (PBG).
The bandgap ∆v = v2 - v1 increases linearly with the index difference ∆n = n2 - n1.
It is clear that the 1D crystal exhibits a photonic bandgap for light propagation along z. It
should also be apparent that there would be no PBGs for propagation along the x and y direction
along which there are no periodic n-variations. Since we can resolve any k-vector along x, y, and
z directions, overall there is no net PBG for light propagation in a 1D PC. The PBG for a 1D
crystal is called a pseudo PBG.
The above ideas can be readily extended to 2D and 3D periodic structures. Again, in principle, there is no full PBG in the 2D PC. In the case of 3D periodic structures, there would be
photonic bandgaps along x, y, and z directions, and for difference polarizations of the electric
field. If the refractive index contrast and the periodicity in the 3D structure are such that these
­photonic bandgaps overlap in all directions and for all polarizations of light, as schematically
depicted in Figure 1.52, the overlap frequency range ∆v becomes a full photonic bandgap in
all directions for all polarizations of light—no light can propagate through this structure over this
95
96
Chapter 1 • Wave Nature of Light
Figure 1.52 (a) The photonic bandgaps along x, y, and z overlap for all polarizations of the field, which results in a
full photonic bandgap ∆v (an intuitive illustration). (b) The unit cell of a woodpile photonic crystal. There are 4 layers,
labeled 1–4 in the figure, with each later having parallel “rods.” The layers are at right angles to each other. Notice
that layer 3 is shifted with respect to 1, and 4 with respect to 2. (c) An SEM image of a woodpile photonic crystal
based on polycrystalline Si; the rod-to-rod pitch d is on the micron scale. (Courtesy of Sandia National Laboratories.)
(d) The optical reflectance of a woodpile photonic crystal showing a photonic bandgap between 1.5 om and 2 om.
The photonic crystal is similar to that in (c) with five layers and d ≈ 0.65 om. (Source: The reflectance spectrum
was plotted using the data appearing in Fig. 3 in S-Y. Lin and J.G. Fleming, J. Light Wave Technol., 17, 1944, 1999.)
frequency range, ∆v. The structure then has a full photonic bandgap. As it turns out not any 3D
periodic structure results in a full photonic bandgap. Only certain 3D periodic structures allow full
photonic bandgaps to develop. One such structure is the wood pile periodic structure shown in Figure
1.52 (b). The unit cell has four layers of rods. The rods are parallel in each layer and the layers are
An SEM image of a 3D photonic crystal
made from porous silicon in which the lattice
structure is close to being simple cubic. The
silicon squares, the unit cells, are connected
at the edges to produce a cubic lattice. This
3D PC has a photonic bandgap centered at
5 om and about 1.9 om wide. (Courtesy
of Max-Planck Institute for Microstructure
Physics.)
An SEM image of a 3D photonic crystal that is based
on the wood pile structure. The rods are polycrystalline
silicon. Although five layers are shown, the unit cell
has four layers, e.g., the four layers starting from the
bottom layer. Typical ­dimensions are in microns. In one
similar structure with rod-to-rod pitch d = 0.65 om
with only a few layers, the Sandia researchers were
able to produce a photonic bandgap ∆l of 0.8 om centered around 1.6 om within the telecommunications
band. (Courtesy of Sandia National Laboratories.)
1.17 • Photonic Crystals
at 90° to each other. An SEM image of a woodpile photonic crystal is shown in Figure 1.52 (c).
The optical reflectance, that is, 1 – transmittance, of this woodpile PC is shown in Figure 1.52 (d).
It is apparent that there is a photonic bandgap, a stop band, over a range of frequencies, or wavelengths, ∆l, around 3 om. The width and the location of the reflectance or the transmittance band
depends on the structure of the photonic crystal, that is, the periodicity, unit cell structure and
refractive index contrast. The colors of certain butterflies and insects arise not from pigments or
colorants but from a photonic crystal effect with the right periodicity.
An important practical aspect of PCs, analogous to semiconductor crystals, is the importance
of defects. Point and line defects that occur in normal crystals also occur in PCs, as illustrated in
Figure 1.53. They are intentionally introduced to endow the PC structure with certain optical properties. A defect is a discontinuity in the periodicity of the PC. For example, if we upset the p­ eriodicity
by removing a unit cell, we create a so-called point defect. This void can act as an optical cavity,
trapping the EM radiation within the cavity as illustrated in Figure 1.53. We can, of course, remove
a group of unit cells, or modify the refractive index over a few unit cells, which would create an
optical microcavity. Defects introduce localized electromagnetic modes with frequencies within the
photonic bandgap. Defects can tightly confine a mode of light in a very small cavity volume. We can
also enhance the refractive index locally, which would also classify as a point defect.
Line defects are formed when a long row of unit cells are missing, or the refractive index
stays constant over a long line in the crystal. Such a line in which the index is constant allows
propagating EM modes within the photonic bandgap. Since EM waves can propagate within and
along the line defect, the line defect acts as an optical wave guide, guiding the radiation along
its length as shown in Figure 1.53. The EM wave cannot spread into the perfect photonic crystal
region since, in this region, the frequency falls into the stop band.
One very important property of photonic crystals is their ability to suppress or inhibit
spontaneous emission. We can understand this effect intuitively by considering a 3D PC made
from a semiconductor. In spontaneous emission, an electron falls from the conduction band
Figure 1.53 Schematic illustration of point and line defects in a photonic crystal. A point defect acts as an
optical cavity, trapping the radiation. Line defects allow the light to propagate along the defect line. The light
is prevented from dispersing into the bulk of the crystal since the structure has a full photonic bandgap. The
frequency of the propagating light is in the bandgap, that is, in the stop band.
97
98
Chapter 1 • Wave Nature of Light
Sajeev John, at the University of Toronto, along with Eli Yablonovitch
(shown earlier in the chapter) carried out the initial pioneering work in the
development of the field of photonics crystals. Sajeev John was able to
show that it is ­possible to trap light in a similar way the electron is captured,
that is localized, by a trap in a semiconductor. Defects in photonic crystals
can confine or localize electromagnetic waves; such ­effects have important
applications in quantum computing and integrated photonics. (Courtesy
of Sajeev John.)
to the valence band spontaneously and emits a photon of energy hv that corresponds to the
bandgap energy Eg However, if the photon frequency hv falls into the bandgap of the PC, then
this photon is not allowed to propagate or “exist” in the structure. It is prevented from being
emitted—the photons have no place to go.
Photonic crystals have also been shown to exhibit a so-called superprism effect. Under
­appropriate conditions, the dispersion of light by a prism-shaped photonic crystal is considerably
enhanced over that corresponding to a homogeneous prism having the same average refractive
index as the photonic crystal. This originates from the strong curvature of the v9k curve near the
edge of the Brillouin Zone as apparent in Figure 1.51 (a). The latter may also be viewed as a high
refractivity variation with wavelength.
Questions and Problems
1.1 Maxwell’s wave equation and plane waves
(a) Consider a traveling sinusoidal wave of the form Ex = Eo cos (vt - kz + fo). The latter can also be
written as Ex = Eo cos 3k(vt - z) + fo4 , where v = v>k is the velocity. Show that this wave satisfies
­Maxwell’s wave equation, and show that v = 1>(moeoer)1>2.
(b) Consider a traveling function of any shape, even a very short delta pulse, of the form Ex = f 3k(vt - z) 4 ,
where f is any function, which can be written is Ex = f(f), f = k(vt - z). Show that this traveling function
satisfies Maxwell’s wave equation. What is its velocity? What determines the form of the function f ?
1.2 Propagation in a medium of finite small conductivity An electromagnetic wave in an isotropic medium
with a dielectric constant er and a finite conductivity s and traveling along z obeys the following equation for
the variation of the electric field E perpendicular to z
d 2E
dz2
- eoermo
02E
0t 2
= mos
0E
0t
(P1.1)
Show that one possible solution is a plane wave whose amplitude decays exponentially with propagation along z,
that is, E = Eo exp ( - a′z) exp 3 j(vt - kz)4 . Here exp ( - a′z) causes the envelope of the amplitude to decay
with z (attenuation) and exp 3j(vt – kz)4 is the traveling wave portion. Show that in a medium in which a is small,
the wave velocity and the attenuation coefficient of the field are given by
v =
v
1
=
k
1moeoer
and a′ =
s
2eocn
where n is the refractive index (n = e1>2
r ). What is the attenuation coefficient a that describes the decay of the
light intensity? (Metals with high conductivities are excluded.)
1.3 Point light source What is the irradiance measured at a distance of 1 m and 2 m from a 1 W light point source?
1.4 Gaussian beam Estimate the divergence and Rayleigh range of a Gaussian beam from a He-Ne Laser with
l = 633 nm and a beam width of 1.00 nm at z = 0. After traversing 10 m through vacuum, what will the
beam width be?
1.5 Gaussian beam in a cavity with spherical mirrors Consider an optical cavity formed by two aligned spherical mirrors facing each other as shown in Figure 1.54. Such an optical cavity is called a spherical mirror
Questions and Problems
resonator, and is most commonly used in gas lasers. Sometimes, one of the ­reflectors is a plane mirror. The two
spherical mirrors and the space between them form an optical resonator because only certain light waves with
certain frequencies can exist in this optical cavity. The radiation inside a spherical mirror cavity is a Gaussian
beam. The actual or particular Gaussian beam that fits into the cavity is that beam whose wavefronts at the mirrors match the curvature of the mirrors. Consider the symmetric resonator shown in Figure 1.54 in which the
mirrors have the same radius of curvature R. When a wave starts at A, its wavefront is the same as the curvature
of A. In the middle of the cavity it has the minimum width and at B the wave again has the same curvature as B.
Such a wave in the cavity can replicate itself (and hence exist in the cavity) as it travels between the mirrors
provided that it has right beam characteristics, that is the right curvature at the mirrors. The radius of curvature
R of a Gaussian beam wavefront at a distance z along its axis is given by
R(z) = z 31 + (zo >z)24 ;
zo = pw2o >l is the Rayleigh range
Figure 1.54 Two spherical mirrors
reflect waves to and from each other.
The optical cavity contains a Gaussian
beam. This particular optical cavity is
symmetric and confocal; the two focal
points coincide at F.
Consider a confocal symmetric optical cavity in which the mirrors are separated by L = R.
(a) Show that the cavity length L is 2zo, that is, it is the same as twice the Rayleigh range, which is the reason
the latter is called the confocal length.
(b) Show that the waist of the beam 2wo is fully determined only by the radius of curvature R of the mirrors,
and given by
2wo = (2lR>p)1>2
1.6
1.7
1.8
1.9
(c) If the cavity length L = R = 50 cm, and l = 633 nm, what is the waist of the beam at the center and also
at the mirrors?
Cauchy dispersion equation Using the Cauchy coefficients and the general Cauchy equation, calculate
refractive index of a silicon crystal at wavelengths of 200 om and at 2 om, over two orders of magnitude wavelength change. What is your conclusion?
Sellmeier dispersion equation Using the Sellmeier equation and the coefficients, obtain a graph of the
­refractive index of fused silica (SiO2) versus its wavelength in the range of 500 nm to 1550 nm.
Sellmeier dispersion equation The Sellmeier dispersion coefficient for pure silica (SiO2) and 86.5%
SiO2-13.5% GeO2 are given in Table 1.2. Write a program on your computer or calculator, or use a math software package or even a spreadsheet program (e.g., Excel) to obtain the refractive index n as a function of l from
0.5 om to 1.8 om for both pure silica and 86.5% SiO2-13.5% GeO2. Obtain the group index, Ng, vs. wavelength
for both materials and plot it on the same graph. Find the wavelength at which the material dispersion, defined
as the derivative of the group velocity with respect to the wavelength, becomes zero in each material.
The Cauchy dispersion relation for zinc selenide ZnSe is a II–VI semiconductor and a very useful optical
material used in various applications such as optical windows (especially high power laser windows), lenses, and
prisms. It transmits over 0.50–19 om. n in the 1–11 om range is described by a Cauchy expression of the form
n = 2.4365 +
0.0485
l2
+
0.0061
l4
- 0.0003l2
in which l is in om. What are the n–2, n0, n2 and n4 coefficients? What is ZnSe’s refractive index n and group
index Ng at 5 om?
99
100
Chapter 1 • Wave Nature of Light
1.10 Refractive index, reflection, and the Brewster’s angle
(a) Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the phase velocity?
(b) What is the Brewster angle (the polarization angle up) and the critical angle (uc) for total internal reflection
when the light wave traveling in this silica medium is incident on a silica–air interface. What happens at the
polarization angle?
(c) What is the reflection coefficient and reflectance at normal incidence when the light beam travel­ing in the
silica medium is incident on a silica–air interface?
(d) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is
incident on an air–silica interface? How do these compare with part (c) and what is your conclusion?
1.11 Snell’s law and lateral beam displacement What is the lateral displacement when a laser beam passes through
two glass plates, each of thickness 1 mm, with refractive indices of n1 = 1.570 and n2 = 1.450 respectively, if
the angle of incidence is 45°?
1.12 Snell’s law and lateral beam displacement An engineer wants to design a refractometer (an ­instrument for
measuring the refractive index) using the lateral displacement of light through a glass plate. His initial experiments involve using a plate of thickness L, and measuring the displacement of a laser beam when the angle of
incidence ui is changed, for example, by rotating (tilting) the sample. For ui = 40° he measures a displacement
of 0.60 mm, and when ui = 80° he measures 1.69 mm. Find the refractive index of the plate and its thickness.
(Note: You need to solve a nonlinear equation for n numerically.)
1.13 Snell’s law and prisms Consider the prism shown in Figure 1.55 that has an apex angle a = 60°. The prism
has a refractive index of n and it is in air.
(a) What are Snell’s law at interfaces at A (incidence and transmittance angles of ui and ut) and B (incidence
and transmittance angles of ui′ and ut′)?
(b) Total deflection d = d1 + d2 where d1 = ui - ut and d2 = ut′ - ui′. Now, b + u i= + ut = 180° and
a + b = 180°. Find the deflection of the beam for an incidence angle of 45° for the following three colors
at which n is known: Blue, n = 1.4634 at l = 486.1 nm; yellow, n = 1.4587 at l = 589.2 nm; red,
n = 1.4567 at l = 656.3 nm. What is the separation in distance b­ etween the rays if the rays are projected
on a screen 1 m away.
Figure 1.55 A light
beam is deflected by a prism
through an angle d. The angle
of incidence is ui. The apex
angle of the prism is a.
1.14 Fermat’s principle of least time Fermat’s principle of least time in simple terms states that when light travels
from one point to another it takes a path that has the shortest time. In going from a point A in some medium with
a refractive index n1 to a point B in a neighboring medium with refractive index n2 as in Figure 1.56, the light
path AOB involves refraction at O that satisfies Snell’s law. The time it takes to travel from A to B is minimum
only for the path AOB such that the incidence and refraction angles ui and ut satisfy Snell’s law. Let’s draw a
straight line from A to B cutting the x-axes at O′. The line AO′B will be our reference line and we will place the
origin of x and y coordinates at O′. Without invoking Snell’s law, we will vary point O along the x-axis (hence
OO′ is a variable labeled x), until the time it takes to travel AOB is minimum, and thereby derive Snell’s law.
The time t it takes for light to travel from A to B through O is
t =
3(x1 - x)2 + y214 1>2 3(x2 + x)2 + y224 1>2
AO
OB
+
=
+
c>n1
c>n2
c>n1
c>n2
(P1.2)
Questions and Problems
Figure 1.56 Consider a light wave
traveling from point A (x1, y2) to B (x1, y2)
through an arbitrary point O at a distance x
from O′. The principle of least time from A to
B requires that O is such that the incidence
and refraction angles obey Snell’s law.
The incidence and transmittance angles are given by
sin ui =
x1 - x
3(x1
- x)2 + y21 4 1>2
and sin ui =
(x2 + x)
3(x2
+ x)2 + y22 4 1>2
(P1.3)
Differentiate Eq. (P1.2) with respect to x to find the condition for the “least time” and then use Eq. (P1.3) in this
condition to derive Snell’s law.
Pierre de Fermat (1601–1665) was
a French mathematician who made
many significant contributions to
modern calculus, number theory,
analytical geometry, and probability. (Courtesy of Mary Evans
Picture Library/Alamy.)
1.15 Antireflection (AR) coating
(a) A laser beam of wavelength 1550 nm from air is launched to a single mode optical fiber with a core refractive index n1 = 1.45. Estimate the refractive index and thickness of film required for an anti-reflecting
coating on this fiber.
(b) A Ge photodiode is designed to operate at 1550 nm, and it is required to have AR coatings to minimize
reflected light. Two possible materials are available for AR coating: SiO2 with a refractive index of 1.46,
and TiO2 with a refractive index of 2.2. Which would be better suited? What would be the thickness for
the AR coating on this photodiode? The refractive index of Ge is about 4.
(c) Consider a Ge photodiode that is designed for operation around 1200 nm. What are the best AR coating
refractive index and thickness if the refractive index of Ge is about 4.0?
1.16 Single- and double-layer antireflection V-coating For a single-layer AR coating of index n2 on a material with index n3( 7 n2 7 n1), as shown in Figure 1.57 (a), the minimum reflectance at normal incidence is
given by
Rmin = c
n22 - n1n3
n22
+ n1n3
d
2
when the reflections A, B, . . . all interfere as destructively as possible. Rmin = 0 when n2 = (n1n3)1>2. The choice
of materials may not always be the best for a single-layer AR coating. Double-layer AR coatings, as shown in
101
102
Chapter 1 • Wave Nature of Light
Figure 1.57 (a) A single-layer AR coating. (b) A double-layer AR coating and (c) its V-shaped
reflectance spectrum over a wavelength range.
Figure 1.57 (b), can achieve lower and sharper reflectance at a specified wavelength as in Figure 1.57 (c).
To reduce the reflection of light at the n1– n4 interface, two layers n2 and n3, each quarter wavelength in the layer
(l>n2 and l>n3) are interfaced between n1 and n4. The reflections A, B, and C for normal incidence result in a
minimum reflectance given by
Rmin = c
n23n1 - n4n22
n23n1
+
n4n22
d
2
The double-layer reflectance vs. wavelength behavior usually has a V-shape, and such coatings are called V-coatings.
(a) Show that double-layer reflectance vanishes when
(n2 >n3)2 = n1 >n4
(b) Consider an InGaAs, a semiconductor crystal with an index 3.8, for use in a photodetector. What is the
reflectance without any AR coating?
(c) What is the reflectance when InGaAs is coated with a thin AR layer of Si3N4? Which material in Table 1.3
would be ideal as an AR coating?
Table 1.3 Typical AR materials and their approximate refractive indices
over the visible wavelengths
n
MgF2
SiO2
Al2O3
CeF3
Sb2O3
Si3N4
SiO
ZrO2
ZnS
TiO2
CdS
1.38
1.46
1.65
1.65
1.9–2.1
1.95
2.0
2.05
2.35
2.35
2.60
(d) What two materials would you choose to obtain a V-coating? Note: The choice of an AR coating also
depends on the technology involved in depositing the AR coating and its effects on the interface states
between the AR layer and the semiconductor. Si1 - xNx is a common AR coating on devices inasmuch as it is
a good passive dielectric layer, its deposition technology is well established and changing its composition
(x) changes its index.
1.17 Single-, double-, and triple-layer antireflection coatings Figure 1.58 shows the reflectance of an uncoated
glass, and glass that has a single- (1), double- (2) and triple- (3) layer AR coatings? The coating details are in
the figure caption. Each layer in single- and double-layer AR coatings has a thickness of l>4, where l is the
wavelength in the layer. The triple-layer AR layer has three coatings with thicknesses l>4, l>2, and l>4. Can
you qualitatively explain the results by using interference? What applications would need single-, double-, and
triple-layer coatings?
Questions and Problems
Figure 1.58 Reflectance vs. wavelength for a glass plate, n = 1.52, with and without AR coatings.
(1) Single-layer AR coating is a quarter wavelength (l>4) thick MgF2, n = 1.38. (2) Double-layer
coating is l>4 thick MgF2 and l>4 thick Al2O3, n = 1.69. (3) Triple-layer coating is l>4 thick MgF2,
l>2 thick ZrO2, n = 2.05, and a l>4 thick CeF3, n = 1.64. (Source: Plotted from data appearing in
Figure 2.2 in S. Chattopadhyay et al., Mater. Sci. Engin. R, 69, 1, 2010.)
1.18 Reflection at glass–glass and air–glass interfaces A ray of light that is traveling in a glass medium of refractive index n1 = 1.460 becomes incident on a less dense glass medium of refractive index n2 = 1.430. Suppose
that the free-space wavelength of the light ray is 850 nm.
(a) What should the minimum incidence angle for TIR be?
(b) What is the phase change in the reflected wave when the angle of incidence ui = 85° and when ui = 90°?
(c) What is the penetration depth of the evanescent wave into medium 2 when ui = 85° and when ui = 90°?
(d) What is the reflection coefficient and reflectance at normal incidence (ui = 0°) when the light beam traveling in the glass medium (n = 1.460) is incident on a glass–air interface?
(e) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air
is incident on an air–glass (n = 1.460) interface? How do these compare with part (d) and what is your
conclusion?
1.19 Dielectric mirror A dielectric mirror is made up of a quarter wave layer of GaAs with nH = 3.38 and AlAs
with nL = 3.00 at around 1550 nm. The light is incident on the mirror from another semiconductor of refractive
index n0 = 3.40. Find out the number of pairs of layers N needed to get 90% reflectance. Find out the bandwidth of the reflected light.
1.20 TIR and polarization at water–air interface
(a) Given that the refractive index of water is about 1.33, what is the polarization angle for light traveling in air
and reflected from the surface of the water?
(b) Consider a diver in sea pointing a flashlight towards the surface of the water. What is the critical angle for
the light beam to be reflected from the water surface?
1.21 Reflection and transmission at a semiconductor–semiconductor interface A light wave with a wavelength
of 890 nm (free-space wavelength) that is propagating in GaAs becomes incident on AlGaAs. The refractive
index of GaAs is 3.60, that of AlGaAs is 3.30.
(a) Consider normal incidence. What are the reflection and transmission coefficients and the reflectance and
transmittance? (From GaAs into AlGaAs.)
(b) What is the Brewster angle (the polarization angle up) and the critical angle (uc) for total ­internal reflection
for the wave in (a); the wave that is traveling in GaAs and incident on the ­GaAs–­AlGaAs interface?
(c) What is the reflection coefficient and the phase change in the reflected wave when the angle of incidence
ui = 79°?
(d) What is the penetration depth of the evanescent wave into medium 2 when ui = 79° and when ui = 89°?
What is your conclusion?
1.22 Phase changes on TIR Consider a light wave of wavelength 870 nm traveling in a semiconductor medium
(GaAs) of refractive index 3.60. It is incident on a different semiconductor medium (AlGaAs) of refractive index
103
104
Chapter 1 • Wave Nature of Light
3.40, and the angle of incidence is 80°. Will this result in total internal reflection? Calculate the phase change in the
parallel and perpendicular components of the reflected electric field.
1.23 Fresnel’s equations Fresnel’s equations are sometimes given as follows:
r# =
r// =
t# =
Ero,#
Eio,#
Ero, //
Eio, //
Eto,#
Eio,#
=
=
=
n1 cos ui - n2 cos ut
n1cos ui + n2 cos ut
n1 cos ut - n2 cos ui
n1 cos ut + n2 cos ui
2n1 cos ui
n1 cos ui + n2 cos ut
and
t// =
Eto,//
Eio,//
=
2n1 cos ui
n1 cos ut + n2 cos ui
Show that these reduce to Fresnel’s equation given in Eqs. (1.6.6) and (1.6.7).
Using Fresnel’s equations, find the reflection and transmission coefficients for normal incidence and
show that
r# + 1 = t# and r// + nt// = 1
where n = n2 >n1.
1.24 Fresnel’s equations Consider a light wave traveling in a glass medium with an index n1 = 1.440 and it is
incident on the glass–air interface. Using Fresnel’s equations only, that is, Eqs. (1.6.6a) and (1.6.6b), calculate the reflection coefficients r› and r// and hence reflectances R› and R// for (a) ui = 25° and (b) ui = 50°.
In the case of ui = 50°, find the phase change f# and f// from the reflection coefficients by writing
r = r exp ( - jf). Compare f# and f// from r› and r// calculations with those calculated from Eqs. (1.6.11)
and (1.6.12).
1.25 Goos-Haenchen phase shift A ray of light which is traveling in a glass medium (1) of refractive index
n1 = 1.460 becomes incident on a less dense glass medium (2) of refractive index n2 = 1.430. Suppose that
the free-space wavelength of the light ray is 850 nm. The angle of incidence ui = 85°. Estimate the lateral
Goos-Haenchen shift in the reflected wave for the perpendicular field component. Recalculate the GoosHaenchen shift if the second medium has n2 = 1 (air). What is your conclusion? Assume that the virtual
reflection occurs from a virtual plane in medium B at a distance d that is roughly the same as the penetration depth. Note that d actually depends on the polarization, the direction of the field, but we will ignore this
dependence.
1.26 Evanescent wave Total internal reflection of a plane wave from a boundary between a more dense medium
(1) n1 and a less dense medium (2) n2 is accompanied by an evanescent wave propagating in medium 2 near
the boundary. Find the functional form of this wave and discuss how its magnitude varies with the distance into
medium 2.
1.27 TIR and FTIR
(a) By considering the electric field component in medium B in Figure 1.21, explain how you can adjust the
amount of transmitted light through a thin layer between two higher refractive index media.
(b) What is the critical angle at the hypotenuse face of a beam splitter cube made of glass with n1 = 1.6 and
having a thin film of liquid with n2 = 1.3. Can you use 45° prisms with normal incidence?
(c) Explain how a light beam can propagate along a layer of material between two different media as shown in
Figure 1.59 (a). Explain what the requirements are for the indices n1, n2, n3. Will there be any losses at the
reflections?
(d) Consider the prism coupler arrangement in Figure 1.59 (b). Explain how this arrangement works for coupling an external light beam from a laser into a thin layer on the surface of a glass substrate. Light is then
propagated inside the thin layer along the surface of the substrate. What is the purpose of the adjustable
coupling gap?
Questions and Problems
Figure 1.59 (a) Light propagation along an optical guide. (b) Coupling of laser light into a thin
layer—optical guide—using a prism. The light propagates along the thin layer.
1.28 Complex refractive index and dielectric constant The complex refractive index N = n - jK can be
defined in terms of the complex relative permittivity er = er1 - jer2 as
= (er1 - jer2)1>2
N = n - jK = e1>2
r
Show that
n = c
(e2r1 + e2r2)1>2 + er1 1>2
(e2r1 + e2r2)1>2 - er1 1>2
d
and K = c
d
2
2
1.29 Complex refractive index Spectroscopic ellipsometry measurements on a germanium crystal at a photon
energy of 1.5 eV show that the real and imaginary parts of the complex relative permittivity are 21.56 and
2.772, respectively. Find the complex refractive index. What is the reflectance and absorption coefficient at
this wavelength? How do your calculations match with the experimental values of n = 4.653 and K = 0.298,
R = 0.419 and a = 4.53 * 106 m-1?
1.30 Complex refractive index Figure 1.26 shows the infrared extinction coefficient K of CdTe. Calculate the
­absorption coefficient a and the reflectance R of CdTe at 60 om and 80 om.
1.31 Refractive index and attenuation in the infrared region—Reststrahlen absorption Figure 1.26 shows the
refractive index n and the extinction coefficient K as a function of wavelength l in the infrared for a CdTe crystal
due to lattice absorption, called Reststrahlen absorption. It results from the ionic polarization of the crystal induced
by the optical field in the light wave. The relative permittivity er due to positive (Cd2+) and negative (Te2-) ions
being made to oscillate by the optical field about their equilibrium positions is given in its simplest form by
er = er= - je″r = erH +
erH - erL
g v
v 2
a b - 1 + j
a b
vT
vT vT
(P1.4)
where erL and erH are the relative permittivity at low (L) and high (H) frequencies, well below and above the infrared peak, g is a loss coefficient characterizing the rate of energy transfer from the EM wave to lattice vibrations
(phonons), and vT is a transverse optical lattice vibration frequency that is related to the nature of bonding between
the ions in the crystal. Table 1.4 provides some typical values for CdTe and GaAs. Equation (P1.4) can be used to
obtain a reasonable approximation to the infrared refractive index n and extinction coefficient K due to Reststrahlen
absorption. (a) Consider CdTe, and plot n and K vs. l from 40 om to 90 om and compare with the experimental
results in Figure 1.26 in terms of the peak positions and the width of the extinction coefficient peak. (b) Consider
Table 1.4
Ionic polarization resonance parameters for CdTe and GaAs
ErL
ErH
VT ( rad s-1 )
G ( rad s-1 )
CdTe
10.20
7.10
2.68 * 1013
0.124 * 1013
GaAs
13.0
13
0.045 * 1013
11.0
5.07 * 10
105
106
Chapter 1 • Wave Nature of Light
GaAs, and plot n and K vs. l from 30 om to 50 om. (c) Calculate n and K for GaAs at l = 38.02 om and compare
with the experimental values n = 7.55 and K = 0.629. (You might want to use a logarithmic scale for K.)
1.32 Coherence length A narrow band pass filter transmits wavelengths in the range 5000 { 0.5 A°. If this filter
is placed in front of a source of white light, what is the coherence length of the transmitted light?
1.33 Spectral widths and coherence
(a) Suppose that frequency spectrum of a radiation emitted from a source has a central frequency yo and a spectral width ∆y. The spectrum of this radiation in terms of wavelength will have a central wavelength lo and a
spectral width ∆l. Clearly, lo = c>yo. Since ∆l V lo and ∆y V yo, using l = c>y, show that the line
width ∆l and hence the coherence length lc are
∆l = ∆y
1.34
1.35
1.36
1.37
1.38
1.39
lo
yo
= ∆y
l2o
c
and lc = c∆t =
l2o
∆l
(b) Calculate ∆l for a lasing emission from a He-Ne laser that has lo = 632.8 nm and ∆y ≈ 1.5 GHz. Find its
coherence time and length.
Coherence lengths Find the coherence length of the following light sources:
(a) An LED emitting at 1550 nm with a spectral width 150 nm;
(b) A semiconductor laser diode emitting at 1550 nm with a spectral width 3 nm;
(c) A quantum well semiconductor laser diode emitting at 1550 nm with a spectral width of 0.1 nm;
(d) A multimode He-Ne laser with a spectral frequency width of 1.5 GHz;
(e) A specially designed single mode and stabilized He-Ne laser with a spectral width of 100 MHz.
Fabry–Perot optical cavity Consider an optical cavity formed between two identical mirrors, each with
­reflectance = 0.97. The refractive index of the medium enclosed between the mirrors is 1. Find out the minimum
length of the optical cavity which can resolve spectral lines of a sodium lamp with line width ∆l = 0.6 nm and
∆l = 589.3 nm. Further, estimate the mode separation in frequency and wavelength. What are the finesse F
and Q factors for this cavity?
Fabry–Perot optical cavity from a ruby crystal Consider a ruby crystal of diameter 1 cm and length 10 cm.
The refractive index is 1.78. The ends have been silvered and the reflectances are 0.99 and 0.95 each. What is
the nearest mode number that corresponds to a radiation of wavelength 694.3 nm? What is the actual wavelength of the mode closest to 694.3 nm? What is the mode separation in frequency and wavelength? What are
the finesse F and Q factor for the cavity?
Fabry–Perot optical cavity spectral width Consider an optical cavity of length 40 cm. Assume the refractive
index is 1, and use Eq. (1.11.3) to plot the peak closest to 632.8 nm for 4 values of R = 0.99, 0.90, 0.75 and 0.6.
For each case find the spectral width dlm, the finesse F and Q. How ­accurate is Eq. (1.11.5) in predicting dlm.
(You may want to use a graphing software for this problem.)
Diffraction A collimated beam of light of wavelength 632.8 nm is incident on a circular aperture of 250 om. Find
out the divergence of the transmitted beam. Obtain the diameter of the transmitted beam at a distance of 10 m. What
would be the divergence if the aperture is a single slit of width 250 om?
Diffraction intensity Consider diffraction from a uniformly illuminated circular aperture of dia­meter D. The
far field diffraction pattern is given by a Bessel function of the first kind and first order, J1, and the intensity at a
point P on the angle ui with respect to the central axis through the aperture is
I(g) = Ioa
2J1(g)
g
b
2
where Io is the maximum intensity, g = (1>2)kD sin u is a variable quantity that represents the ­angular position
u on the screen as well as the wavelength (k = 2p>l) and the aperture diameter D. J1(g) can be calculated from
p
J1(g) =
1
cos (a - g sin a)da
pL
0
where a is an integration variable. Using numerical integration, or a suitable mathematics software program,
plot 3J1(g)>g4 vs. g for g = 0 - 8 and confirm that zero-crossings occur at g = 3.83, 7.02 and the maxima at
g = 0, 5.14. What is the intensity ratio of the first bright ring (at g = 5.14) to that at the center of the Airy disk
(g = 0)? (You can use a very small g instead of zero for the center intensity calculation.) Using the first zero at
g = 3.83, verify Eq. (1.12.5), sin uo = 1.22l>D, where uo is the angular position of the first dark ring, as defined
in Figure 1.36 (b).
Questions and Problems
George Bidell Airy (1801–1892, England).
George Airy was a ­professor of astronomy at
Cambridge and then the Astronomer Royal at
the Royal Observatory in Greenwich, England.
(© Mary Evans Picture Library/Alamy.)
1.40 Bragg diffraction A reflection grating is made on the surface of a semiconductor with a periodicity of 0.5 om.
If light of wavelength 1.55 om is incident at an angle of 88° to the normal, find out the diffracted beam.
1.41 Diffraction grating for WDM Consider a transmission diffraction grating. Suppose that we wish to use this
grating to separate out different wavelengths of information in a WDM signal at 1550 nm. (WDM stands of
wavelength division multiplexing.) Suppose that the diffraction grating has a periodicity of 2 om. The angle of
incidence is 0° with respect to the normal to the diffraction grating. What is the angular separation of the two
wavelength component s at 1.550 om and 1.540 om? How would you increase this separation?
1.42 A monochromator Consider an incident beam on a reflection diffraction grating as in Figure 1.60. Each incident wavelength will result in a diffracted wave with a different diffraction angle. We can place a small slit
and allow only one diffracted wave lm to pass through to the photodetector. The diffracted beam would consist of wavelengths in the incident beam separated (or fanned) out after diffraction. Only one wavelength lm
will be diffracted favorably to pass through the slit and reach the photodetector. Suppose that the slit width is
s = 0.1 mm, and the slit is at a distance R = 5 cm from the grating. Suppose that the slit is placed so that it is
at right angles to the incident beam: ui + um = p>2. The grating has a corrugation periodicity of 1 om.
Figure 1.60 A mono­
chromator based on using a
diffraction grating.
(a) What is the range of wavelengths that can be captured by the photodetector when we rotate the grating from
ui = 1° to 40°?
(b) Suppose that ui = 15°. What is the wavelength that will be detected? What is the resolution, that is, the
range of wavelengths that will pass through the slit? How can you improve the resolution? What would be
the advantage and disadvantage in decreasing the slit width s?
107
108
Chapter 1 • Wave Nature of Light
1.43 Thin film optics Consider light incident on a thin film on a substrate, and assume normal incidence for
simplicity.
(a) Consider a thin soap film in air, n1 = n3 = 1, n2 = 1.40. If the soap thickness d = 1 om, plot the
reflectance vs. wavelength from 0.35 om to 0.75 om, which includes the visible range. What is your
conclusion?
(b) MgF2 thin films are used on glass plates for the reduction of glare. Given that n1 = 1, n2 = 1.38, and
n3 = 1.60 (n for glass depends on the type of glass but 1.6 is a reasonable value), plot the reflectance as
a function of wavelength from 0.35 om to 0.75 om for a thin film of thickness 0.10 om. What is your
conclusion?
1.44 Thin film optics Consider a glass substrate with n3 = 165 that has been coated with a transparent optical film (a dielectric film) with n2 = 2.50, n1 = 1 (air). If the film thickness is 500 nm, find the minimum
and maximum reflectances and transmittances and their corresponding wavelengths in the visible range
for normal incidence. (Assume normal incidence.) Note that the thin n2-film is not an AR coating, and for
n1 6 n3 6 n2,
R max = a
n22 - n1n3
n22
+ n1n3
b
2
and R min = a
n3 - n1
n3 + n1
b
2
1.45 Thin film optics Consider light incident on a thin film on a substrate, and assume normal incidence for simplicity. Plot the reflectance R and transmittance T as a function of the phase change f from f = - 4p to + 4p
for the following cases
(a) Thin soap film in air, n1 = n3 = 1, n2 = 1.40. If the soap thickness d = 1 om, what are the maxima and
minima in the reflectance in the visible range?
(b) A thin film of MgF2 on a glass plate for the reduction of glare, where n1 = 1, n2 = 1.38, and n3 = 1.70
(n for glass depends on the type of glass but 1.7 is a reasonable value.) What should be the thickness of
MgF2 for minimum reflection at 550 nm?
(c) A thin film of semiconductor on glass where n1 = 1, n2 = 3.5, and n3 = 1.55.
1.46 Transmission through a plate Consider the transmittance of light through a partially transparent glass plate
of index n in which light experiences attenuation (either by absorption or scattering). Suppose that the plate is in
a medium of index no, the reflectance at each n–no interface is R and the attenuation coefficient is a.
(a) Show that
Tplate =
(1 - R)2e-ad
(1 - R2)e-2ad
(b) If T is the transmittance of a glass plate of refractive index n in a medium of index no show that, in the
absence of any absorption in the glass plate,
n>no = T -1 + (T -2 - 1)1>2
if we neglect any losses in the glass plate.
(c) If the transmittance of a glass plate in air has been measured to be 89.96%. What is its refractive index? Do
you think this is a good way to measure the refractive index?
1.47 Scattering Consider Rayleigh scattering. If the incident light is unpolarized, the intensity Is of the scattered
light a point at a distance r at an angle u to the original light beam is given by
Is ∝
1 - cos2 u
r2
Plot a polar plot of the intensity Is at a fixed distance r from the scatter as we change the angle u around the
scatterer. In a polar plot, the radial coordinate (OP in Figure 1.48 (b)) is Is. Construct a contour plot in the xy
plane in which a contour represents a constant intensity. You need to vary r and u or x and y such that Is remains
constant. Note x = r cos u and y = r sin u, u = arctan (y>x), r = (x2 + y2)1>2.
1.48 One-dimensional photonic crystal (a Bragg mirror) The 1D photonic crystal in Figure 1.50 (a), which is
essentially a Bragg reflector, has the dispersion behavior shown in Figure 1.51 (a). The stop-band ∆v for normal incidence and for all polarizations of light is given by (R. H. Lipson and C. Lu, Eur. J. Phys., 30, S33, 2009)
n2 - n1
∆v
= (4>p) arcsin a
b
vo
n2 + n1
Questions and Problems
where ∆v is the stop band, vo is the center frequency defined in Figure 1.51 and n2 and n1 are the high and low
refractive indices. Calculate the lowest stop band in terms of photon energy in eV, and wavelength 1550 nm
for a 1D photonic crystal structure with n1d1 = n2d2 = l>4, made up of: (i) Si (nSi = 3.5) and SiO2
(nSiO2 = 1.445) pairs, and (ii) Si3N4 (nSi3N4 = 2.0) and SiO2 pairs.
1.49 Photonic crystals Concepts have been borrowed from crystallography, such as a unit cell, to ­define a photonic crystal. What is the difference between a unit cell used in a photonic crystal and that used in a real crystal?
What is the size limit on the unit cell of a photonic crystal? Is the refractive index a microscopic or a macroscopic concept? What is the assumption on the refractive index?
A scanning Fabry–Perot interferometer (Model SA200), used as a spectrum analyzer, which has a free spectral range of 1.5 GHz, a typical finesse of 250, spectral width (resolution) of 7.5 MHz. The cavity length is
5 cm. It uses two concave mirrors instead of two planar mirrors to form the optical cavity. A piezoelectric
transducer is used to change the cavity length and hence the resonant frequencies. A voltage ramp is applied
through the coaxial cable to the piezoelectric transducer to scan frequencies. (Courtesy of Thorlabs.)
This is a tunable large aperture (80 mm) etalon with two end plates that act as reflectors. The end plates
have been ­machined to be flat to wavelength >110. There are three piezoelectric transducers that can tilt
the end plates and hence obtain ­perfect alignment. (Courtesy of Light Machinery.)
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$IBQUFS t 4FNJDPOEVDUPS 4DJFODF BOE -JHIU&NJUUJOH %JPEFT
Questions and Problems
3.1 Metals and work function The metal sodium (Na) has an atomic concentration of 2.54 * 1022 cm-3. Each
Na atom in isolation has one outer valence electron in an unfilled 3s-subshell. Once the solid is formed, the outer
valence electrons from all Na atoms are shared in the crystal by all Na+ ions, that is, these valence electrons find
themselves in an unfilled energy band as in Figure 3.1 (c). Since each Na atom donates one electron to the ­energy
band, the electron concentration is the same as the atomic concentration. Calculate the Fermi energy at 0 K.
What is the speed of the electrons at EF? What should be the speed at 300 K if the electrons could be treated
classically in terms of the kinetic molecular theory as if they were free, similar to the atoms in a gas?
3.2 Photocathode and work function The photocathode of a photomultiplier tube has a multi-alkaline (Sb-NaK-Cs) metal with a work function (Φ) of 1.55 eV. What is the longest wavelength that will cause photoemission? What is the kinetic energy of a photoemitted electron if the incident light wavelength is 450 nm (blue)?
The quantum efficiency (QE) of a photocathode is defined by
Quantum efficiency =
Number of photoemitted electrons
Number of incident photons
The QE is 100% if each incident photon ejects one electron. Suppose that blue light of wavelength 450 nm with
an intensity of 1 oW cm-2 is incident on this photocathode with an area of 50 mm2. If the emitted electrons are
collected by applying a positive bias voltage to an anode, and the photocathode has a QE of 25%, what will be
the photocurrent? (Normally the photoemitted electron is accelerated by a suitable applied field and impacts
­another electrode, a dynode, where it causes secondary electron emission, and so on, until the current is multiplied by orders of magnitude.)
3.3 Refractive index and bandgap Diamond, silicon, and germanium all have the same diamond unit cell. All
three are covalently bonded solids. Their refractive indices (n) and energy bandgaps (Eg) are shown in Table
3.2. (a) Plot n vs. Eg and (b) Plot also n4 vs. 1 >Eg. What is your conclusion? According to Moss’s rule, roughly,
n4Eg ≈ K, a constant. What is the value of K?
TAbLE 3.2 T
he refractive index n and the bandgap Eg of diamond,
Ge, and Si, all of which have the same crystal structure
Material S
Bandgap, Eg (eV)
n
Diamond
Silicon
Germanium
5
2.4
1.1
3.46
0.66
4.0
3.4 Electrons in the CB of a nondegenerate semiconductor
(a) Consider the energy distribution of electrons nE(E) in the conduction band. Assuming that the density of
state gCB(E) ∝ (E - Ec)1>2 and using Boltzmann statistics f (E) ≈ exp 3- (E - EF)>kBT4, show that the
energy distribution of the electrons in the CB can be written as
nx(x) = Cx1>2 exp ( - x)
where x = (E - Ec)>kBT is the electron energy in terms of kBT measured from Ec and C is a constant at a
given temperature (independent of E).
(b) Setting arbitrarily C = 1, plot nx(x) vs. x. Where is the maximum and what is the FWHM (full width at
half maximum, that is, between half maximum points)? Is the use of 1.8kBT for the half-maximum width
correct?
(c) Show that the average electron energy in the CB is (3>2) kBT, by using the definition of average,
∞
xaverage =
∞
xnxdx> nxdx
L0
L0
where the integration is from x = 0 (Ec) to say x = 10 (far away from Ec where nx S 0). You need to use
a numerical integration.
(d) Show that the maximum in the energy distribution is at x = (1>2) or at Emax = (1>2) kBT.
2VFTUJPOT BOE 1SPCMFNT
3.5 Intrinsic and doped GaAs The properties of GaAs are shown in Table 3.1. Calculate the intrinsic concentration and the intrinsic resistivity at room temperature (take as 300 K). Where is the Fermi level? Assuming the
Nc and Nv scale as T 3>2, what would be the intrinsic concentration at 100°C? If this GaAs crystal is doped with
1017 donors cm-3 (such as Te), where is the new Fermi level and what is the resistivity of the sample? The drift
motilities in GaAs are shown in Table 3.3.
TAbLE 3.3 Ionized dopant impurities scatter carriers and reduce the drift mobility.
The dependence of Me for electrons and Mh for holes on the total ionized
dopant concentration
Dopant concentration (cm−3 )
GaAs, me (cm2 V-1 s-1)
GaAs, mh (cm2 V-1 s-1)
Si, me (cm2 V-1 s-1)
Si, mh (cm2 V-1 s-1)
0
1014
1015
1016
1017
1018
8500
400
1450
490
–
–
1420
485
8000
380
1370
478
7000
310
1200
444
5000
250
730
328
2400
160
280
157
3.6 Electrons in GaAs Given that the electron effective mass m *e for the GaAs is 0.067me, calculate the thermal
velocity of the electrons in the CB of a nondegenerately doped GaAs at room temperature (300 K). If me is
the drift mobility of the electrons and te the mean free time between electron scattering events (between elec2 -1 -1
trons and lattice vibrations) and if me = ete >m*,
e calculate te, given me = 8500 cm V s . Calculate the drift
­velocity vd = me E of the CB electrons in an applied field E of 105 V m-1. What is your conclusion?
3.7 Extrinsic n-GaAs An n-type GaAs crystal is doped with 1016 donors cm-3 (such as Te), what are the electron
and hole concentrations, and the conductivity? (See Table 3.3.)
3.8 Extrinsic n-Si A Si crystal has been doped n-type with 1 * 1017 cm-3 phosphorus (P) donors. The electron
drift mobility me depends on the total concentration of ionized dopants Ndopant, as in Table 3.3, inasmuch as
these ionized dopants scatter the electrons and thereby decrease their drift mobility. What is the conductivity of
the sample? Where is the Fermi level with respect to the intrinsic crystal?
3.9 Compensation doping in n-type Si An n-type Si sample has been doped with 1016 phosphorus (P) atoms cm-3.
(a) What are the electron and hole concentrations? (b) Calculate the room temperature conductivity of the sample. (c) Where is the Fermi level with respect to EFi? (d) If we now dope the crystal with 1017 boron acceptors,
what will be the electron and hole concentrations? (e) Where is the Fermi level with respect to EFi?
3.10 Free carrier absorption in semiconductors The attenuation of light due to the optical field drifting the free carriers is called free carrier absorption. As the free electrons in a semiconductor crystal are accelerated by the optical
field, they eventually become scattered by lattice vibrations or impurities, and pass the energy absorbed from the
radiation to lattice vibrations. In such cases, er″ and the AC conductivity s at the same frequency are related by
er″ = s>eov
(P3.1)
We consider a semiconductor in which the free carriers are electrons (an n-type semiconductor). The AC conductivity s in general is given by
s = so >(1 + jvte)
(P3.2)
where so is the DC conductivity, v is the angular frequency of light, and te is the scattering time of the conduction electrons. s decreases with frequency in Eq. (P3.2). Consider an n-type semiconductor. The free carriers
are the electrons in the CB. If the drift mobility of the electrons is me, then me = ete >m*,
e where m*
e is the
­effective mass of the electrons in the CB of the semiconductor.28 The DC conductivity so = enme, where n is
the concentration of CB electrons. Show that the absorption coefficient due to free carrier absorption (due to the
conductivity) when v 7 1>t is given by
a = a
so
1
e3n
ba b = a 2 3
b l2
nceot2 v2
4p nc eom*e 2me
(P3.3)
28
As we know, conduction in an n-type semiconductor occurs by the drift of free electrons inside the semiconductor crystal. The dopants donate electrons to the crystal, which are free within the crystal.
271
272
$IBQUFS t 4FNJDPOEVDUPS 4DJFODF BOE -JHIU&NJUUJOH %JPEFT
where n is the refractive index. What would you expect if you plotted a vs. l2? Consider a Si crystal doped with
1015 cm-3 donors. Estimate the free carrier absorption (in m-1 and dB m-1) at 1.55 and at 5 om. What is your
conclusion? [Although Eq. (P3.3) is a highly simplified approximation to describing free carrier absorption, it
nonetheless provides a rough estimate of its magnitude.]
3.11 GaAs pn junction Consider a GaAs pn junction that has the following properties: Na = 1016 cm-3 (p-side);
Nd = 1018 cm-3 (n-side); B = 2.0 * 10-16 m3 s-1; cross-sectional area A = 1.5 mm * 1.5 mm. Assume a
long diode. What is the diode current due to diffusion in the neutral regions and recombination in the SCL at 300 K
when the forward voltage across the diode is 0.8 V and then 1.1 V? (Use the drift motilities in Table 3.3 for
calculating the diffusion coefficients through the Einstein relation.)
3.12 InP pn junction Consider an InP pn junction that has the following properties: Na = 1015 cm-3 (p-side);
Nd = 1017 cm-3 (n-side); using B ≈ 4 * 10-16 m3 s-1; cross-sectional area A = 1 mm * 1 mm. Assume a
long diode. What is the diode current due to diffusion in the neutral regions and recombination in the SCL at
300 K when the forward voltage across the diode is 0.70 and 0.9 V? The electron mobility in the p-side is about
∼6000 cm2 s-1 and the hole mobility on the n-side is roughly ∼100 cm2 s-1. (See also Table 3.1 for ni and er.)
Comment on the ideality factor of this InP pn junction.
3.13 Si pn junction Consider a long pn junction diode with an acceptor doping Na of 1018 cm-3 on the p-side and
donor concentration of Nd on the n-side. The diode is forward biased and has a voltage of 0.6 V across it. The
diode cross-sectional area is 1 mm2. The minority carrier recombination time, t, depends on the total dopant
concentration, Ndopant (cm-3), through the following approximate empirical relation
t ≈ (5 * 10-7)>(1 + 2 * 10-17Ndopant)
where t is in seconds.
(a) Suppose that Nd = 1015 cm-3. Then the depletion layer extends essentially into the n-side and we have to
consider minority carrier recombination time, th, in this region. Calculate the diffusion and recombination
contributions to the total diode current given Na = 1018 cm-3 and Nd = 1015 cm-3. Use Table 3.3 for me
and mh. What is your conclusion?
(b) Suppose that Nd = Na. Then W extends equally to both sides and, further, te = th. Calculate the diffusion and recombination contributions to the diode current given Na = 1018 cm-3 and Nd = 1018 cm-3. Use
Table 3.3 for me and mh. What is your conclusion?
3.14 Injected minority carrier charge Consider a pn junction with heavier doping on the p-side. The injected
minority carriers (holes) represent an injected excess minority carrier charge Qh in the neutral region as shown
in Figure 3.24 (a). (There is also excess majority carrier charge so the region is neutral.) Show that
Q = Ith for a long diode and Q = Itt for a short diode
in which th is the hole lifetime and tt is the diffusion time, or the transit time of holes across the width of the
neutral n-region, that is, tt = l 2n >2Dh. What is your conclusion?
3.15 High injection condition The Shockley equation for a pn junction under forward bias, as shown in Figure
3.16 (a), was derived by assuming low (weak) injection conditions, that is pn(0) ≈ ∆pn(0) V nno or Nd on the
n-side. Show that when the injection is no longer weak, that is when pn(0) ≈ nno = Nd, the applied voltage V
reaches VSI (strong injection) given by
VSI = Vo - Vth ln (Na >Nd)
where Vth is the thermal voltage (kBT>e). Calculate Vo and VSI for a Si pn junction that has Na = 1018 cm-3 and
Nd = 1016 cm-3. Can you use the Shockley equation when V 7 VSI? What happens when Na = Nd ? What is
your conclusion?
3.16 Heterostructure Consider a Type I heterostructure as shown in Figure 3.27.
(a) If Eg1 6 Eg2 and if x1 and x2 are the electron affinities of each semicondcutor, show that
∆Ec = x1 - x2 and
∆Ev = Eg2 - Eg1 - ∆Ec
(b) Using the data in Table 3.1, draw the energy band diagram of an nP junction between an n-type Ge and
P-GaAs. Under forward bias, is it easier to inject electrons or holes?
(c) Draw the energy band diagram for a pN junction between p-type Ge and N-GaAs. Under forward bias, is it
easier to inject electrons or holes?
3.17 Heterojunction I–V characteristics We use some of the data reported by Womac and Rediker (J. Appl.
Phys., 43, 4130, 1972) for AlGaAs>GaAs pn heterojunction at 298 K (25°C). Sample A is an N + p and sample
B is a P + n junction. The I–V data of interest are listed in Table 3.4. By a suitable plot find the ideality factor for
each. What is your conclusion?
2VFTUJPOT BOE 1SPCMFNT
273
TAbLE 3.4 I–V data on two heterojunctions. First set is N + p and the second set is a P + n junction
A: Heterojunction N + p
V 0.206 V 0.244 0.290
I 1.03 nA 2.07
5.20
0.322
10.3
0.362
20.7
0.412
52.8
0.453
105
0.485
192
0.537
515
B: Heterojunction P + n
V 0.310 V 0.364 0.402
I 2.01 nA 4.91
9.79
0.433
19.0
0.485
49.5
0.521
96.1
0.561
194
0.608
466
0.682
1.96 o A
0.576
1.02 oA
0.726
5.02
0.612
2.03
0.662
4.89
0.708
10.1
0.764
9.75
0.807
19.5
0.859
50.6
3.18 AlGaAs LED emitter An AlGaAs LED emitter for in a local optical fiber network has the output spectrum
shown in Figure 3.32 (b). It is designed for peak emission at about 822 nm at 25°C. (a) Why does the peak emission wavelength increase with temperature? (b) What is the bandgap of AlGaAs in this LED? (c) The bandgap,
Eg, of the ternary AlxGa 1 - xAs alloys follows the empirical expression, Eg(eV) = 1.424 + 1.266x + 0.266x2.
What is the composition of the AlxGa 1 - xAs in this LED?
3.19 III–V compound semiconductors in optoelectronics Figure 3.52 represents the bandgap Eg and the lattice
­parameter a in a quaternary III–V alloy system. A line joining two points represents the changes in Eg and a with
composition in a ternary alloy composed of the compounds at the ends of that line. For example, starting at GaAs
point, Eg = 1.42 eV and a = 0.565 nm, Eg decreases and a increases as GaAs is alloyed with InAs, as we move
FIGUrE 3.52 Bandgap energy
Eg and lattice constant a for various
III–V alloys of GaP, GaAs, InP, and
InAs. A line represents a ternary
alloy formed with compounds from
the end points of the line. Solid
lines are for direct bandgap alloys
whereas dashed lines for indirect
bandgap alloys. Regions between lines
represent quaternary alloys. The line
from X to InP represents quaternary
alloys In1 - xGa xAs1 - yPy made from
In0.53Ga0.47As and InP, which are
lattice-matched to InP.
InGaAsP 1300 nm LED emitters, each pigtailed to an optical
fiber for use in optical communication modems and lower speed
data>analog transmission systems. (Courtesy of OSI Laser
Diode, Inc.)
0.885
99.5
274
$IBQUFS t 4FNJDPOEVDUPS 4DJFODF BOE -JHIU&NJUUJOH %JPEFT
along the line joining GaAs to InAs. Eventually at InAs, Eg = 0.35 eV and a = 0.606 nm. Point X in Figure 3.52
is composed of InAs and GaAs and it is the ternary alloy In1 - xGa xAs. At X, In0.53Ga0.47As (often called “in-gas”
in telecom) has Eg = 0.73 eV and a = 0.587 nm, which is the same a as that for InP. In1 - xGa xAs at X is therefore lattice-matched to InP and can hence be grown on an InP substrate without creating defects at the interface.
Further, In1 - xGa x As at X can be alloyed with InP to obtain a quaternary alloy29 In1 - xGa xAs1 - yPy whose
properties lie on the line joining X and InP and therefore all have the same lattice parameter as InP but different
bandgap. Layers of In1 - xGa xAs1 - yPy with composition between X and InP can be grown epitaxially on an InP
substrate by various techniques such as liquid phase epitaxy (LPE) or molecular beam epitaxy (MBE). The grey
shaded area between the solid lines represents the possible values of Eg and a for the quaternary III–V ­alloy
system in which the bandgap is direct and hence suitable for direct recombination. The compositions of the
quaternary alloy lattice matched to InP follow the line from X to InP.
(a) Given that the In1 - xGa x As at X is In0.53Ga0.47As show that quaternary alloys In1 - xGa xAs1 - yPy are lattice
matched to InP when y = 1 - 2.13x.
(b) The bandgap energy Eg, in eV for In1 - xGa xAs1 - yPy lattice-matched to InP is given by the empirical relation, Eg (eV) = 0.75 + 0.46y + 0.14 y2. Find the composition of the quaternary alloy suitable for an LED
emitter operating at 1.30 om.
3.20 Varshni equation and the change in the bandgap with temperature The Varshni equation describes the
change in the energy bandgap Eg of a semiconductor with temperature T as given by Eq. (3.11.2) that is
Eg = Ego - AT 2 >(B + T)
where Ego is Eg at 0 K, and A and B are constants. Show that
dEg
dT
= -
AT(T + 2B)
(B + T)
2
= -
(Ego - Eg) T + 2B
a
b
T
T + B
For GaAs, Ego = 1.519 eV, A = 5.41 * 10-4 eV K-1, B = 204 K. What is dEg >dT for GaAs? Find the shift
in the emitted wavelength from a GaAs LED per 1°C change at room temperature (300 K). Find the emission
wavelength at 27°C and - 30°C.
3.21 Dependence on the emission peak and linewidth on temperature Using the Varshni equation find the
peak emission wavelength and the linewidth of the emission spectrum from an In0.47Ga0.53As LED when it is
cooled from 25°C to - 25°C. You can use Eqs. (3.3.1) and (3.3.2). The Varshni constants for In0.47Ga0.53As are
Ego = 0.814 eV, A = 4.906 * 10-4 eV K-1, B = 301 K.
3.22 LED Output Spectrum Given that the width of the relative light intensity vs. photon energy spectrum of an
LED is typically around ∼3 kBT, calculate the spectral width in wavelength (nm) of LED emitters operating at
850 nm, 1310 nm, and 1550 nm?
3.23 Linewidth of LEDs Experiments carried out on various direct bandgap semiconductor LEDs give the output
spectral linewidth (between half intensity points as listed in Table 3.5). From Figure 3.31 we know that a spread
in the wavelength is related to a spread in the photon energy, ∆l ≈ (hc>E 2ph)∆Eph where Eph = hy is the photon energy. Suppose that we write Eph = hc>l and ∆Eph = ∆(hy) ≈ mkBT where m is a numerical constant.
Therefore,
∆l ≈ (mkBT>hc)l2
(P3.4)
By appropriately plotting the data in Table 3.5, and assuming T = 300 K, find m.
TAbLE 3.5 L
inewidth 𝚫L1,2 between half points in the output spectrum (spectral
intensity vs. wavelength) of eight LEDs using direct bandgap semiconductors
29
Material (Direct Eg)
Peak wavelength of
emission (l) nm
AlGaAs
AlGaAs
AlGaAs
GaAs
GaAs
InGaAsP
InGaAsP
InGaAsP
650
810
820
890
950
1150
1270
1500
∆l1>2 nm
22
36
40
50
55
90
110
150
Some books have other formats for the chemical composition for example, Ga xIn1 - xAsyP1 - y. The present notation
In1 - xGa xAs1 - yPy was chosen to reflect the common vernacular for InGaAs (pronounced “in-gas”).
Physical Constants
c
h
Speed of light in vacuum
Planck’s constant
h
h = h>2p
e
eo
kB
kBT>e
me
mo
NA
R
Electronic charge
Absolute permittivity
Boltzmann constant (kB = R>NA)
Thermal voltage at 300 K
Electron mass in free space
Absolute permeability
Avogadro’s number
Gas constant (NAkB)
Useful Information
p = 3.1416
°
1 A (Angstrom) = 0.1 nm = 10-10 m
2.9979 * 108 m s-1
6.6261 * 10-34 J s
4.1357 * 10-15 eV s
1.0546 * 10-34 J s
6.5821 * 10-16 eV s
1.60218 * 10-19 C
8.8542 * 10-12 F m-1
1.3807 * 10-23 J K-1
0.02585 V
9.10939 * 10-31 kg
4p * 10-7 H m-1
6.0221 * 1023 mol-1
8.31457 J mol-1 K-1
e = 2.7183
1 eV = 1.60218 * 10-19 J
Common Prefixes for Multiples of Ten
10–15
f
femto
10–12
p
pico
10–9
n
nano
10–6
o
micro
10–3
m
milli
10–2
c
centi
103
k
kilo
106
M
mega
109
G
giga
1012
T
tera